The figure shows a network of resistors and a | vedclass

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

Question

(B) Let the potential at point $H$ be $0\,V$. Then the potential at $A$ is $E$. Branch $CF$ has a resistance of $2\,\Omega$ and a current of $1\,A$ fl

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) Let the potential at point $H$ be $0\,V$. Then the potential at $A$ is $E$. Branch $CF$ has a resistance of $2\,\Omega$ and a current of $1\,A$ flows through it. Thus,the potential difference across $CF$ is $V_C - V_F = 1\,A 	imes 2\,\Omega = 2\,V$. Since $F$ is connected to $H$ (ground),$V_F = 0\,V$,so $V_C = 2\,V$. Now,consider the branch $CD-DE$. The resistance is $1\,\Omega + 1\,\Omega = 2\,\Omega$. The current through this branch is $I_{CD} = (V_C - V_E) / 2\,\Omega = (2\,V - 0\,V) / 2\,\Omega = 1\,A$. The total current flowing through $BC$ is $I_{BC} = I_{CF} + I_{CD} = 1\,A + 1\,A = 2\,A$. The potential at $B$ is $V_B = V_C + I_{BC} 	imes 2\,\Omega = 2\,V + (2\,A 	imes 2\,\Omega) = 6\,V$. The current through branch $BG$ is $I_{BG} = V_B / 6\,\Omega = 6\,V / 6\,\Omega = 1\,A$. The total current flowing from the battery is $I = I_{BC} + I_{BG} = 2\,A + 1\,A = 3\,A$. The potential at $A$ is $V_A = V_B + I 	imes 2\,\Omega = 6\,V + (3\,A 	imes 2\,\Omega) = 12\,V$. Since $V_A = E$,the $emf$ of the battery is $12\,V$.

The figure shows a network of resistors and a battery. If $1\,A$ current flows through the branch $CF$,then find the $emf$ $E$ of the battery. (in $,V$)

Similar Questions

The magnitude and direction of the current in the circuit shown will be

When three identical bulbs of $60 \;W, 200 \;V$ rating are connected in series to a $200 \;V$ supply,the power drawn by them will be ....... $Watt$.

Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered.

In the circuit shown, the heat produced in $5 \Omega$ resistance due to current $I_2$ is $50 \text{ J/s}$. Then, the heat generated per second in $2 \Omega$ resistance is (in $\text{ J/s}$)

$A$ set of $n$ equal resistors,each of value $R$,are connected in series to a battery of emf $E$ and internal resistance $R$. The current drawn is $I$. Now,the $n$ resistors are connected in parallel to the same battery. The current drawn from the battery becomes $10I$. The value of $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module