Question

A battery of emf $10V$ and internal resistance $3\Omega$ is connected to a resistor. The current in the circuit is $0.5A$. The terminal voltage of the battery when the circuit is closed is?

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Solution

Step 1: Given DataEMF of the battery $\epsilon =10V$The internal resistance of the battery $r=3\Omega$The current in the circuit $I=0.5A$Let the resistance of the resistor be $R$.Let the terminal voltage be $V$.Step 2: Formula Used$I=\frac{\epsilon }{R+r}$Step 3: Calculate the ResistanceBy Ohm’s law, we know that$I=\frac{\epsilon }{R+r}\phantom{\rule{0ex}{0ex}}⇒IR+Ir=\epsilon \phantom{\rule{0ex}{0ex}}⇒R=\frac{\epsilon -Ir}{I}$Upon substituting the values we get,$R=\frac{10-\left(0.5×3\right)}{0.5}\phantom{\rule{0ex}{0ex}}=\frac{10-1.5}{0.5}\phantom{\rule{0ex}{0ex}}=17\Omega$Step 4: Calculate the Terminal VoltageAccording to Ohm's law, we get,$V=I×R\phantom{\rule{0ex}{0ex}}=0.5×17\phantom{\rule{0ex}{0ex}}=8.5V$Hence, the required terminal voltage is $8.5V$.

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