KCET · Physics · Current Electricity
A galvanometer of resistance \(240 \Omega\) allows only \(4 \%\) of the main current after connecting a shunt resistance. The value of the shunt resistance is
- A \(10 \Omega\)
- B \(20 \Omega\)
- C \(8 \Omega\)
- D \(5 \Omega\)
Answer & Solution
Correct Answer
(A) \(10 \Omega\)
Step-by-step Solution
Detailed explanation
Given, galvanometer resistance \(G=240 \Omega\)
Shunt resistance \(S=\) ?
\(\mathrm{I}_{\mathrm{G}}=\frac{4}{100} \mathrm{I}\)
From figure voltage through the circuit.
\(\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) \mathrm{S}=\mathrm{I}_{\mathrm{G}} \mathrm{G}\)
or \(\quad\left(1-\frac{4 \mathrm{I}}{100}\right) \mathrm{S}=\frac{4 \mathrm{I}}{100} \times 240\)
or
\(S=\frac{4 \times 240}{96}=10 \Omega\)
Shunt resistance \(S=\) ?
\(\mathrm{I}_{\mathrm{G}}=\frac{4}{100} \mathrm{I}\)
From figure voltage through the circuit.
\(\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) \mathrm{S}=\mathrm{I}_{\mathrm{G}} \mathrm{G}\)
or \(\quad\left(1-\frac{4 \mathrm{I}}{100}\right) \mathrm{S}=\frac{4 \mathrm{I}}{100} \times 240\)
or
\(S=\frac{4 \times 240}{96}=10 \Omega\)
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