Two cells of e.m.f. \(E_1\) and \(E_2\) (\(E_1 > E_2\)) are connected as shown in figure.
When a potentiometer is connected between points A and B the balancing length of potentiometer wire is \(412\) cm. When same potentiometer wire is connected between points A and C the balancing length is \(103\) cm. The ratio \(E_1:E_2\) is
Show Hint
The cells oppose each other, so A to C measures E1 minus E2.
Step 1: Understanding the Figure:
The circuit is a chain A, cell \(E_1\), point B, cell \(E_2\), point C. In the symbols, the longer line is the positive terminal. \(E_1\) has its positive side toward A, and \(E_2\) has its positive side toward C, so the two cells are connected in opposition at B.
Step 2: Potential differences:
Across AB only \(E_1\) acts, so the balancing length is \(412\) cm for \(E_1\). Across AC the net emf is \(E_1 - E_2\), balanced at \(103\) cm.
Step 3: Use proportionality to length:
\(\frac{E_1}{E_1 - E_2} = \frac{412}{103} = 4\). So \(E_1 = 4E_1 - 4E_2\), giving \(4E_2 = 3E_1\).
Step 4: Ratio:
\[ E_1 : E_2 = 4 : 3 \]
Step 5: Why the other options are wrong.
6:1 and 4:1 would result from assuming \(E_1 + E_2\) or \(E_2\) alone is balanced. 3:4 is the inverse, which would violate \(E_1 > E_2\).
Final Answer:
The ratio \(E_1 : E_2\) is 4:3, option (C).
\[ \boxed{4:3} \]
