The ratio of powers dissipated respectively i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the current through resistor $R$ be $I_1$ and the current through resistor $2R$ be $I_2$. Since $R$ and $2R$ are connected in parallel,the pot

Vedclass · Accepted Answer

$4/27$

Vedclass · Answer

(D) Let the current through resistor $R$ be $I_1$ and the current through resistor $2R$ be $I_2$. Since $R$ and $2R$ are connected in parallel,the potential difference across them is the same,so $V_{AB} = V_{CD}$.Therefore,$I_1 R = I_2 (2R)$,which gives $I_2 = I_1 / 2$.The total current flowing through the $3R$ resistor is $I_3 = I_1 + I_2 = I_1 + I_1 / 2 = (3/2) I_1$.The power dissipated in resistor $R$ is $P_1 = I_1^2 R$.The power dissipated in resistor $3R$ is $P_3 = I_3^2 (3R) = ((3/2) I_1)^2 (3R) = (9/4) I_1^2 (3R) = (27/4) I_1^2 R$.The ratio of power dissipated in $R$ to that in $3R$ is $\frac{P_1}{P_3} = \frac{I_1^2 R}{(27/4) I_1^2 R} = \frac{4}{27}$.

The ratio of powers dissipated respectively in $R$ and $3R$,as shown in the figure,is:

Similar Questions

For the circuits $A$ and $B$ as shown in the figure,identify the correct option.

Resistances $R_1$ and $R_2$,each $60\,\Omega$,are connected in series as shown in the figure. The potential difference between $A$ and $B$ is maintained at $120\,V$. What will be the reading of the voltmeter connected between points $C$ and $D$ if the resistance of the voltmeter is $120\,\Omega$?

What is immaterial for an electric fuse wire?

What is the current in $A$ flowing through the given circuit?

Two batteries of e.m.f $4 \text{ V}$ and $8 \text{ V}$ with internal resistance $1 \Omega$ and $2 \Omega$ respectively are connected in a circuit with a resistance of $9 \Omega$ as shown in the figure. The current and potential difference between the points $P$ and $Q$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module