Electrical Energy and Power Questions in English

(B) The power $P$ delivered to the device is given by $P = V I$,where $V$ is the voltage and $I$ is the current flowing through the wires.
From this,the current $I$ can be expressed as $I = \frac{P}{V}$.
The power wasted in the connecting wires due to their finite resistance $R_C$ is given by the formula $P_C = I^2 R_C$.
Substituting the expression for $I$ into the power loss formula,we get $P_C = \left( \frac{P}{V} \right)^2 R_C$.
Therefore,the power wasted is $P_C = \frac{P^2 R_C}{V^2}$.

(B) The current $I$ splits into $I_1$ and $I_2$ through the parallel resistors $R$ and $2R$. Using the current divider rule:
$I_1 = I \times \frac{2R}{R + 2R} = \frac{2I}{3}$
$I_2 = I \times \frac{R}{R + 2R} = \frac{I}{3}$
The heat produced in a resistor is given by $H = I^2 R t$. Assuming time $t$ is the same for all:
$H_1 = I_1^2 R = \left(\frac{2I}{3}\right)^2 R = \frac{4I^2 R}{9}$
$H_2 = I_2^2 (2R) = \left(\frac{I}{3}\right)^2 (2R) = \frac{2I^2 R}{9}$
For the resistor $1.5R$,the total current $I$ flows through it:
$H_3 = I^2 (1.5R) = 1.5 I^2 R = \frac{13.5 I^2 R}{9}$
The ratio $H_1 : H_2 : H_3 = \frac{4}{9} : \frac{2}{9} : \frac{13.5}{9} = 4 : 2 : 13.5 = 8 : 4 : 27$.

(A) The work done by a battery of emf $\varepsilon$ to drive a charge $q$ through the circuit is given by $W = \varepsilon q$.
Since the current $I$ flows for time $t$,the total charge transferred is $q = I t$.
Therefore,the work done by the battery is $W = \varepsilon I t$.
Note: The options provided in the question are potentially confusing. The total work done by the battery is $\varepsilon I t$. The energy dissipated as heat in the external resistor $R$ is $I^2 R t$,and in the internal resistor $r$ is $I^2 r t$. The total energy is $I^2(R+r)t = \varepsilon I t$. Given the standard form of such questions,if the question asks for the work done by the battery,the correct expression is $\varepsilon I t$.

(D) When a charge $\Delta q$ moves through a potential difference $V$,the change in its electric potential energy is given by $\Delta U = \Delta q \cdot V$. Since $I = \frac{\Delta q}{\Delta t}$,we have $\Delta q = I \Delta t$. Thus,the potential energy decreases by $I V \Delta t$.
In a steady current,the drift velocity of the charge carriers remains constant,meaning the kinetic energy of the charge does not increase significantly. Instead,the lost potential energy is dissipated as heat due to collisions with the lattice ions of the conductor.
Therefore,the thermal energy of the conductor increases by $I V \Delta t$.
Statements $II$ and $III$ are correct.

(B) The power consumed by a bulb is given by $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the constant resistance of the bulb.
Taking the logarithmic derivative, we get $\frac{dP}{P} = 2 \frac{dV}{V}$.
Here, the percentage change in voltage is $\frac{dV}{V} = -2.5 \% = -0.025$.
Therefore, the percentage change in power is $\frac{dP}{P} = 2 \times (-2.5 \%) = -5 \%$.
The negative sign indicates a decrease in power.
Thus, the power decreases by $5 \%$ of its rated value.

(A) Since the resistors are connected in parallel,the potential difference $(V)$ across each resistor is the same.
Power consumed by a resistor is given by the formula $P = \frac{V^2}{R}$.
Since $V$ is constant for both resistors,we have $P \propto \frac{1}{R}$.
Therefore,the ratio of power consumed is $\frac{P_1}{P_2} = \frac{R_2}{R_1}$.
Given $R_1 = 10 \Omega$ and $R_2 = 20 \Omega$,we get:
$\frac{P_1}{P_2} = \frac{20}{10} = \frac{2}{1}$.
Thus,the ratio of power consumed by them is $2: 1$.

(A) The heat energy $(U)$ produced in a metallic conductor of resistance $(R)$ in a given time $(t)$ is given by the formula:
$U = \frac{V^2}{R} t$
Since the resistance $(R)$ and time $(t)$ are constant,we have:
$U \propto V^2$
This relationship represents a parabola that opens upwards,starting from the origin $(0,0)$.
Therefore,the graph in option $(A)$ correctly represents this variation.

(B) The power $P$ and voltage $V$ are given as $P = 550 \,W$ and $V = 220 \,V$.
The relationship between power, voltage, and current $I$ is given by the formula $P = V \times I$.
Rearranging for current, we get $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{550}{220} = 2.5 \,A$.
Therefore, the current drawn by the heater is $2.5 \,A$.

(B) Given:
Electromotive force (emf) $\varepsilon = 12 \, V$
Total charge capacity $q = 80 \, A \cdot h$
Power output $P = 120 \, W$
The total energy $E$ stored in the battery is given by $E = q \cdot \varepsilon$.
Substituting the values: $E = 80 \, A \cdot h \times 12 \, V = 960 \, W \cdot h$.
Since power $P$ is the rate of energy delivery, $P = E / \Delta t$, where $\Delta t$ is the time.
Therefore, $\Delta t = E / P = 960 \, W \cdot h / 120 \, W = 8 \, h$.
Thus, the battery can deliver energy at the rate of $120 \, W$ for $8 \, h$.

(C) Given: Current $I = 4 \ A$, Voltage $V = 220 \ V$, Mass of water $m = 1 \ kg$, Initial temperature $T_1 = 34^{\circ} C$, Final temperature $T_2 = 100^{\circ} C$.
Specific heat capacity of water $c = 4200 \ J/(kg \cdot ^{\circ} C)$.
Electric power $P = V \times I = 220 \times 4 = 880 \ W$.
Heat energy required $Q = mc\Delta T = 1 \times 4200 \times (100 - 34) = 4200 \times 66 = 277200 \ J$.
Since electric energy is converted into heat energy, $P \times t = Q$.
$880 \times t = 277200$.
$t = 277200 / 880 = 315 \ seconds$.
To convert time into minutes: $t = 315 / 60 = 5.25 \ minutes$.

(B) The power $P$ dissipated in a resistor is given by the formula $P = \frac{V^2}{R}$,where $V$ is the potential difference and $R$ is the resistance.
Given:
Potential difference $V = 120 \text{ V}$
Resistance $R = 100 \Omega$
Substituting the values into the formula:
$P = \frac{(120)^2}{100}$
$P = \frac{14400}{100}$
$P = 144 \text{ W}$
Therefore,the power dissipated is $144 \text{ W}$.

(D) When the lamp is in use,the resistance $R$ is given by the formula $R = \frac{V^2}{P}$.
Substituting the given values: $R = \frac{240 \times 240}{60} = 960 \ \Omega$.
Let $R'$ be the resistance of the cold filament (when not in use).
According to the problem,$R = 20 \times R'$.
Therefore,$R' = \frac{R}{20} = \frac{960}{20} = 48 \ \Omega$.

(A) Given: Rated power,$P_R = 2400 \, W$ and Rated voltage,$V_R = 240 \, V$.
Since the resistance $R$ of the filament is constant,we use the formula $P = \frac{V^2}{R}$.
First,calculate the resistance $R$ of the heating element:
$R = \frac{V_R^2}{P_R} = \frac{240 \times 240}{2400} = 24 \, \Omega$.
Now,when the element is connected to a $V = 120 \, V$ supply,the new power dissipated $P'$ is:
$P' = \frac{V^2}{R} = \frac{120 \times 120}{24} = 600 \, W$.

(A) The power $P$ consumed by a bulb is given by the formula $P = \frac{V^2}{R}$, where $V$ is the voltage and $R$ is the resistance of the filament.
Since the voltage source is the same for all bulbs, $V$ is constant.
Therefore, the relationship between power and resistance is $R = \frac{V^2}{P}$, which implies $R \propto \frac{1}{P}$.
This means that the resistance is inversely proportional to the power rating of the bulb.
To have the highest resistance, the bulb must have the lowest power rating.
Comparing the given powers $(100 \,W, 200 \,W, 500 \,W, 1000 \,W)$, the $100 \,W$ bulb has the lowest power.
Thus, the $100 \,W$ bulb filament has the highest resistance.

(D) Given,the resistance $R$ of a device is related to the current $I$ by the relation: $R = \frac{0.2 I}{I-4}$.
Power $P$ consumed by the device is given by $P = I^2 R$.
Substituting the expression for $R$: $P = I^2 \left( \frac{0.2 I}{I-4} \right) = \frac{0.2 I^3}{I-4}$.
To find the minimum power,we differentiate $P$ with respect to $I$ and set it to zero: $\frac{dP}{dI} = 0$.
$\frac{d}{dI} \left( \frac{0.2 I^3}{I-4} \right) = 0$.
Using the quotient rule: $\frac{(I-4)(0.6 I^2) - (0.2 I^3)(1)}{(I-4)^2} = 0$.
$0.6 I^3 - 2.4 I^2 - 0.2 I^3 = 0$.
$0.4 I^3 - 2.4 I^2 = 0$.
$0.4 I^2 (I - 6) = 0$.
Since $I > 4$,we have $I = 6 \ A$.
Substituting $I = 6 \ A$ into the power equation: $P_{\min} = \frac{0.2 \times (6)^3}{6-4} = \frac{0.2 \times 216}{2} = 0.2 \times 108 = 21.6 \ W$.

(B) The total current $I$ flowing through the wire is given by the integral of current density $J$ over the cross-sectional area $A$: $I = \int J \, dA$.
Since the wire is circular,$dA = 2 \pi r \, dr$.
$I = \int_0^{R} J(r) \cdot 2 \pi r \, dr$,where $R = 2 \times 10^{-3} \text{ m}$.
$I = \int_0^{2 \times 10^{-3}} (10^5 r) \cdot (2 \pi r) \, dr = 2 \pi \times 10^5 \int_0^{2 \times 10^{-3}} r^2 \, dr$.
$I = 2 \pi \times 10^5 \left[ \frac{r^3}{3} \right]_0^{2 \times 10^{-3}} = 2 \pi \times 10^5 \times \frac{8 \times 10^{-9}}{3} = \frac{16 \pi}{3} \times 10^{-4} \text{ A}$.
The energy consumed $E$ is given by $E = V \cdot I \cdot t$.
$E = 70 \text{ V} \times \left( \frac{16 \pi}{3} \times 10^{-4} \text{ A} \right) \times 1000 \text{ s}$.
$E = 70 \times \frac{16 \pi}{3} \times 10^{-1} = \frac{1120 \pi}{30} = \frac{112}{3} \pi \approx 37.33 \pi \text{ J}$.
Given the options,the closest value is $37 \pi \text{ J}$ (Note: The unit in the provided options appears to be in Joules,not kJ).

(B) Given: Current density $J = (2 \times 10^{10}) r^2 \text{ A/m}^2$, potential $V = 50 \text{ V}$, time $t = 100 \text{ s}$, and radius $R = 2 \times 10^{-3} \text{ m}$.
Energy $E$ dissipated as heat is given by $E = VIt$.
First, calculate the total current $I$ using $I = \int J dA$, where $dA = 2 \pi r dr$.
$I = \int_{0}^{R} (2 \times 10^{10} r^2) (2 \pi r dr) = 4 \pi \times 10^{10} \int_{0}^{2 \times 10^{-3}} r^3 dr$.
$I = 4 \pi \times 10^{10} \left[ \frac{r^4}{4} \right]_{0}^{2 \times 10^{-3}} = \pi \times 10^{10} \times (2 \times 10^{-3})^4$.
$I = \pi \times 10^{10} \times 16 \times 10^{-12} = 16 \pi \times 10^{-2} = 0.16 \pi \text{ A}$.
Now, calculate energy $E = VIt = 50 \times (0.16 \pi) \times 100$.
$E = 5000 \times 0.16 \pi = 800 \pi \text{ J}$.

(A) The rate of heat production (power) in a resistor is given by the formula $H = \frac{V^2}{R}$.
In the first case,the power is $H = \frac{V^2}{R}$.
In the second case,the new voltage is $V' = \frac{V}{3}$ and the new resistance is $R' = 2R$.
The new rate of heat production $H'$ is given by:
$H' = \frac{(V')^2}{R'} = \frac{(\frac{V}{3})^2}{2R} = \frac{\frac{V^2}{9}}{2R} = \frac{V^2}{18R}$.
Since $H = \frac{V^2}{R}$,we can substitute this into the equation:
$H' = \frac{H}{18}$.

(C) The resistance $R$ of the bulb is constant and is determined by its rated values:
$R = \frac{V_{rated}^2}{P_{rated}} = \frac{200^2}{50} = \frac{40000}{50} = 800 \, \Omega$
When connected to a new supply voltage $V' = 100 V$,the new power $P'$ consumed by the bulb is given by:
$P' = \frac{(V')^2}{R} = \frac{100^2}{800} = \frac{10000}{800} = 12.5 \, W$
Therefore,the present power of the bulb is $12.5 \, W$.

(D) The power delivered to the factory is $P_{out} = 10 \text{ kW} = 10000 \text{ W}$ at a voltage $V = 200 \text{ V}$.
The current $I$ flowing through the cable is given by $I = \frac{P_{out}}{V} = \frac{10000}{200} = 50 \text{ A}$.
The power loss in the cable due to its resistance $R = 0.2 \Omega$ is $P_{loss} = I^2 R = (50)^2 \times 0.2 = 2500 \times 0.2 = 500 \text{ W}$.
The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 10000 + 500 = 10500 \text{ W}$.
The efficiency of transmission is $\eta = \frac{P_{out}}{P_{in}} \times 100 = \frac{10000}{10500} \times 100 \approx 95.24 \%$.
Rounding to the nearest given option, the efficiency is $95 \%$.

(A) The total charge $Q$ passing through the coil is equal to the area under the current-time graph. Since the current decreases uniformly from $I_0$ to $0$ in time $T$,the graph is a triangle with base $T$ and height $I_0$.
$Q = \frac{1}{2} I_0 T \Rightarrow I_0 = \frac{2 Q}{T}$
The current as a function of time is given by $I(t) = I_0 \left(1 - \frac{t}{T}\right) = \frac{2 Q}{T} \left(1 - \frac{t}{T}\right)$.
The heat generated $H$ in the coil is given by $H = \int_0^T I^2 R \, dt$.
$H = R \int_0^T \left[ \frac{2 Q}{T} \left(1 - \frac{t}{T}\right) \right]^2 \, dt = \frac{4 Q^2 R}{T^2} \int_0^T \left(1 - \frac{t}{T}\right)^2 \, dt$.
Let $u = 1 - \frac{t}{T}$,then $du = -\frac{1}{T} dt$,so $dt = -T du$.
When $t=0, u=1$; when $t=T, u=0$.
$H = \frac{4 Q^2 R}{T^2} \int_1^0 u^2 (-T \, du) = \frac{4 Q^2 R}{T} \int_0^1 u^2 \, du$.
$H = \frac{4 Q^2 R}{T} \left[ \frac{u^3}{3} \right]_0^1 = \frac{4 Q^2 R}{3 T}$.

(D) The current $I$ in the circuit is given by $I = \frac{E}{R+r}$.
The power $P$ consumed by the external resistance $R$ is $P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$.
To find the condition for maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = 0$.
$\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$.
This implies $(R+r)^2 - 2R(R+r) = 0$.
Dividing by $(R+r)$,we get $(R+r) - 2R = 0$,which simplifies to $r - R = 0$ or $R = r$.
Thus,the power consumption is maximum when the external resistance equals the internal resistance of the battery.

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$,Voltage $V = 250 \ V$,Resistance $R = 2 \ \Omega$.
The current $I$ flowing through the line is $I = \frac{P_{out}}{V} = \frac{1000}{250} = 4 \ A$.
The power loss in the transmission line is $P_{loss} = I^2 R = (4)^2 \times 2 = 16 \times 2 = 32 \ W$.
The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 32 = 1032 \ W$.
The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100$.
$\eta = \left( \frac{1000}{1032} \right) \times 100 \approx 96.898 \% \approx 96.9 \%$.

(C) The maximum power is drawn from a battery when the external resistance $R$ is equal to the internal resistance $r$ of the battery $(R = r)$.
The power delivered to the external resistor is given by the formula: $P = I^2 R = \left(\frac{E}{R+r}\right)^2 \cdot R$.
Substituting $R = r$ for maximum power,we get:
$P_{max} = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.
Given values are $E = 6.0 \ V$ and $r = 0.2 \ \Omega$.
Substituting these values into the formula:
$P_{max} = \frac{(6.0)^2}{4 \times 0.2} = \frac{36}{0.8} = 45 \ W$.
Therefore,the maximum power drawn is $45 \ W$.

(C) The resistance $R$ of the heater is constant.
Using the formula $R = \frac{V^2}{P}$, we calculate $R = \frac{220^2}{400} \Omega$.
The new power consumed $P'$ at the new voltage $V' = 200 \text{ V}$ is given by $P' = \frac{(V')^2}{R}$.
Substituting the value of $R$, we get $P' = \frac{(V')^2 \cdot P}{V^2} = \frac{200^2 \times 400}{220^2}$.
$P' = \frac{40000 \times 400}{48400} = \frac{16000000}{48400} \approx 330.57 \text{ W}$.
Rounding off to the nearest integer, we get $331 \text{ W}$.