In the circuit shown,the thermal power dissip | vedclass

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

Question

(B) The current $i$ flows through $R_1 = R$. The power dissipated in $R_1$ is $P = i^2 R$.In the parallel combination of $R_2 = R$ and $R_3 = 2R$,the

Vedclass · Accepted Answer

$\frac{4 P}{9}$

Vedclass · Answer

(B) The current $i$ flows through $R_1 = R$. The power dissipated in $R_1$ is $P = i^2 R$.In the parallel combination of $R_2 = R$ and $R_3 = 2R$,the current $i$ divides. Let $i_2$ be the current through $R_2$ and $i_3$ be the current through $R_3$.By the current divider rule,$i_2 = i 	imes \frac{R_3}{R_2 + R_3} = i 	imes \frac{2R}{R + 2R} = i 	imes \frac{2R}{3R} = \frac{2}{3}i$.The thermal power dissipated in $R_2$ is $P_2 = i_2^2 R_2 = (\frac{2}{3}i)^2 R = \frac{4}{9} i^2 R$.Since $P = i^2 R$,we have $P_2 = \frac{4}{9} P$.

In the circuit shown,the thermal power dissipated in $R_1$ is $P$. The thermal power dissipated in $R_2$ is

Similar Questions

In the circuit shown,the resistance of the voltmeter is $10,000 \, \Omega$ and that of the ammeter is $20 \, \Omega$. The ammeter reading is $0.10 \, A$ and the voltmeter reading is $12 \, V$. Then $R$ is equal to ............. $\Omega$.

The potential difference between the points $P$ and $Q$ is nearly: (in $V$)

$A$ battery with an internal resistance of $4 \Omega$ is connected to a circuit consisting of three resistors,$R$,$2R$,and $4R$,as shown in the figure. If the power generated in the circuit is maximum,then the magnitude of $R$ must be: (in $Omega$)

In the adjoining figure,the $emf$ of the cell is $2 \, V$ and its internal resistance is negligible. The resistance of the voltmeter is $80 \, \Omega$. The reading of the voltmeter will be ................ $volt$.

In the circuit element given here,if the potential at point $B$,$V_B = 0$,then the potentials of $A$ and $D$ are given as

Vedclass Test Series

Exam Paper Generator

Online Exam Module