Two cells P and Q each of emf 2.16 text{ V} a | vedclass

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(B) Let $E = 2.16 	ext{ V}$ be the emf of each cell and $r_1, r_2$ be their internal resistances. The external resistance is $R = 19.6 	ext{ } \Omeg

Vedclass · Accepted Answer

$2$ : $3$

Vedclass · Answer

(B) Let $E = 2.16 	ext{ V}$ be the emf of each cell and $r_1, r_2$ be their internal resistances. The external resistance is $R = 19.6 	ext{ } \Omega$.The current $I$ in the series circuit is given by $I = \frac{E + E}{r_1 + r_2 + R} = \frac{4.32}{r_1 + r_2 + 19.6}$.For cell $P$,the terminal voltage is $V_P = E - I r_1$. Given $V_P = 2 	ext{ V}$,we have $2 = 2.16 - I r_1$,so $I r_1 = 0.16$.For cell $Q$,the terminal voltage is $V_Q = E - I r_2$. Given $V_Q = 1.92 	ext{ V}$,we have $1.92 = 2.16 - I r_2$,so $I r_2 = 0.24$.Taking the ratio of these two equations:$\frac{I r_1}{I r_2} = \frac{0.16}{0.24} = \frac{16}{24} = \frac{2}{3}$.Thus,the ratio of internal resistances is $r_1 : r_2 = 2 : 3$.

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