Two batteries of e.m.f. 4,V and 8,V with inte | vedclass

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Question

(A) The total e.m.f. of the circuit is $E_{eq} = 8\,V - 4\,V = 4\,V$ (since the batteries are connected in opposition).The total resistance of the cir

Vedclass · Accepted Answer

$\frac{1}{3}\,A$ and $3\,V$

Vedclass · Answer

(A) The total e.m.f. of the circuit is $E_{eq} = 8\,V - 4\,V = 4\,V$ (since the batteries are connected in opposition).The total resistance of the circuit is $R_{total} = r_1 + r_2 + R = 1\,\Omega + 2\,\Omega + 9\,\Omega = 12\,\Omega$.Using Ohm's law,the current $i$ in the circuit is $i = \frac{E_{eq}}{R_{total}} = \frac{4\,V}{12\,\Omega} = \frac{1}{3}\,A$.The potential difference between points $P$ and $Q$ is the voltage drop across the $9\,\Omega$ resistor.$V_{PQ} = i 	imes R = \frac{1}{3}\,A 	imes 9\,\Omega = 3\,V$.

Two batteries of e.m.f. $4\,V$ and $8\,V$ with internal resistances $1\,\Omega$ and $2\,\Omega$ 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 $Q$ are

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