A galvanometer of resistance 20\(\Omega\) is to be provided with a shunt that \(\frac{1}{10}\) of the whole current in a circuit passes through the galvanometer. The resistance of the shunt is
A
2.00\(\Omega\)
B
2.22\(\Omega\)
C
18.00\(\Omega\)
D
18.22\(\Omega\)
Verified Correct Answer: B.
Physical Principles & Numerical Solution
Let the total current be \( I \). The galvanometer current is \( I_g = \frac{1}{10}I \), and the shunt current is \( I_s = I - I_g = \frac{9}{10}I \).
Since the galvanometer resistance \( R_g = 20 \Omega\) and shunt (resistance \( R_s \)) are in parallel, the voltage across them is equal:
\( I_g R_g = I_s R_s\)
\(\left( \frac{1}{10}I \right) \cdot 20 = \left( \frac{9}{10}I \right) \cdot R_s\)
Divide through by \( I \) (since \( I \neq 0 \)):
\(2 = \frac{9}{10} R_s\)
Solve for \( R_s \):
\(R_s = 2 \cdot \frac{10}{9} = \frac{20}{9}\) \(\approx\) 2.222\(\Omega\)
Source: Myschool Past Questions Archive (Record #28478)