JEE Mains · Physics · STD 12 - 3. current electricity
In this figure the resistance of the coil of galvanometer \(G\) is \(2\,\Omega\). The emf of the cell is \(4\,V\). The ratio of potential difference across \(C_1\) and \(C_2\) is:
- A \(1\)
- B \(\frac{4}{5}\)
- C \(\frac{3}{4}\)
- D \(\frac{5}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{5}\)
Step-by-step Solution
Detailed explanation
At steady state, current in the circuit is \(i =\frac{4 V }{6+2+8}=\frac{1}{4}\,A\) Voltage across \(C_1\) is \(V _1= V _{ AC }=i(6 \Omega+2 \Omega)=\frac{1}{4} \times 8=2\,V\) Voltage across \(C _2\) is \(V _2= V _{ BD }=i(2 \Omega+8 \Omega)=\frac{1}{4} \times 10=2.5\,V\)…
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