Two batteries of e.m.f 4 text{ V} and 8 text{ | vedclass

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(B) The circuit consists of two batteries connected in series opposition. The net e.m.f is $E_{net} = 8 	ext{ V} - 4 	ext{ V} = 4 	ext{ V}$.The tot

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

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(B) The circuit consists of two batteries connected in series opposition. The net e.m.f is $E_{net} = 8 	ext{ V} - 4 	ext{ V} = 4 	ext{ V}$.The total resistance of the circuit is $R_{total} = R + r_1 + r_2 = 9 \Omega + 1 \Omega + 2 \Omega = 12 \Omega$.The current in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{4 	ext{ V}}{12 \Omega} = \frac{1}{3} 	ext{ A}$.The potential difference between points $P$ and $Q$ is the voltage across the external resistor $R = 9 \Omega$.$V_{PQ} = I 	imes R = \frac{1}{3} 	ext{ A} 	imes 9 \Omega = 3 	ext{ V}$.

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

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