A d.c. main supply of e.m.f. , 220 , V is con | vedclass

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Question

(C) Let the $d.c.$ supply voltage be $V_s = 220 \, V$,the $e.m.f.$ of the battery be $E = 200 \, V$,and the internal resistance be $r = 1 \, \Omega$.

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Let the $d.c.$ supply voltage be $V_s = 220 \, V$,the $e.m.f.$ of the battery be $E = 200 \, V$,and the internal resistance be $r = 1 \, \Omega$. For the battery to be charged,the potential difference across the battery terminals must be greater than its $e.m.f.$ $E$. Let the current flowing from the supply be $I$. The circuit consists of the supply $V_s$ in series with the resistance $r$ and the parallel combination of the battery $E$ and the external resistance $R$. However,the problem implies the battery is being charged by the supply through the series resistance $r$. For charging to occur,the terminal voltage $V_t$ across the battery must satisfy $V_t > E$. Using the circuit loop equation: $V_s = I(r + R_{eq})$,where $R_{eq}$ is the effective resistance. Actually,the condition for charging is that the current $I$ must flow into the positive terminal of the battery. The voltage across the battery terminals is $V_t = V_s - I r$. For charging,$V_t > E \implies V_s - I r > E$. Substituting the values: $220 - I(1) > 200 \implies I Also,the current $I$ is given by $I = \frac{V_s}{r + R} = \frac{220}{1 + R}$. Substituting this into the inequality: $\frac{220}{1 + R} $220  10 \, \Omega$. The minimum value of $R$ is $11 \, \Omega$.

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