In the figure shown,battery 1 has emf = 6, V | vedclass

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(C) The two batteries are connected in series such that their emfs oppose each other. The net emf of the circuit is $E_{net} = E_1 - E_2 = 6\, V - 2\,

Vedclass · Accepted Answer

$5$

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(C) The two batteries are connected in series such that their emfs oppose each other. The net emf of the circuit is $E_{net} = E_1 - E_2 = 6\, V - 2\, V = 4\, V$.The total resistance of the circuit is $R_{total} = r_1 + r_2 = 1\, \Omega + 3\, \Omega = 4\, \Omega$.The current flowing in the circuit is $I = \frac{E_{net}}{R_{total}} = \frac{4\, V}{4\, \Omega} = 1\, A$.Since the current flows into the positive terminal of battery $2$,it is being charged.The potential difference across the terminals of battery $2$ is given by $V_2 = E_2 + I r_2$.Substituting the values,$V_2 = 2\, V + (1\, A 	imes 3\, \Omega) = 2\, V + 3\, V = 5\, V$.

In the figure shown,battery $1$ has $emf = 6\, V$ and internal resistance $= 1\, \Omega$. Battery $2$ has $emf = 2\, V$ and internal resistance $= 3\, \Omega$. The wires have negligible resistance. What is the potential difference across the terminals of battery $2$ (in $, V$)?

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