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 \(8.5\text V\) .
\(E=10\text V\)
\(r=3\Omega \)
\(I=0.5\text A\)
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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