Two cells A and B of emf 2 text{ V} and 1.5 t | vedclass

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(C) The current $i$ flowing through the circuit is given by the net emf divided by the total resistance of the circuit.Since the cells are connected i

Vedclass · Accepted Answer

$V_{A}=1.875 	ext{ V}, V_{B}=1.625 	ext{ V}$

Vedclass · Answer

(C) The current $i$ flowing through the circuit is given by the net emf divided by the total resistance of the circuit.Since the cells are connected in opposition, the net emf is $\varepsilon_{net} = 2 	ext{ V} - 1.5 	ext{ V} = 0.5 	ext{ V}$.The total resistance of the circuit is $R_{total} = 5 	ext{ } \Omega + 5 	ext{ } \Omega + 10 	ext{ } \Omega = 20 	ext{ } \Omega$.Therefore, the current $i = \frac{0.5 	ext{ V}}{20 	ext{ } \Omega} = 0.025 	ext{ A}$.For cell $A$, which is discharging, the terminal potential difference is $V_{A} = \varepsilon_{A} - i r_{A} = 2 	ext{ V} - (0.025 	ext{ A} 	imes 5 	ext{ } \Omega) = 2 - 0.125 = 1.875 	ext{ V}$.For cell $B$, which is charging, the terminal potential difference is $V_{B} = \varepsilon_{B} + i r_{B} = 1.5 	ext{ V} + (0.025 	ext{ A} 	imes 5 	ext{ } \Omega) = 1.5 + 0.125 = 1.625 	ext{ V}$.

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