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A battery of emf  \(10V\) and internal resistance  \(3\Omega \)  is connected to a resistor. If the current in the circuit is  \(0.5A\), what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
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A battery of emf  \(10V\) and internal resistance  \(3\Omega \)  is connected to a resistor. If the current in the circuit is  \(0.5A\), what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer

\(8.5 V\)
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Solution

Here, we are given emf  \((E)\)  as  \(10V,\) Resistance  \((r)\)  as  \(3\Omega \)  and current  \(I\)  is  \(0.5A.\) 
 \(E=10\text V\)
 \(r=3\Omega \)
 \(I=0.5\text A\)
Let the unknown resistance be  \(R\Omega \) .
Using Ohm's law,
 \(V=\mathit{IR}\)
\(E=I(R+r)\)
 \(10=\left(0.5\right)\times (3+R)\)
 \(\frac{10}{0.5}=3+R\)
 \(20=R+3\)
 \(R=17\Omega .\)
Hence, the resistance of the resistor will be  \(17\Omega \) .
To verify the voltage of the battery,
 \(V=\mathit{IR}\)
 \(V=0.5\times 17\)
 \(V=8.5\text V\)
Hence the voltage of the battery is  \(8.5\text V\) .
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