⚡ Solution: Parallel Batteries Analysis (2072 Set D Q. No. 9a)
1. 📋 Clearly Defined Given Parameters
| Component | EMF (\(E\)) | Internal Resistance (\(r\)) |
|---|---|---|
| Battery 1 | \(E_1 = 6\,\text{V}\) | \(r_1 = 0.5\,\Omega\) |
| Battery 2 | \(E_2 = 10\,\text{V}\) | \(r_2 = 1\,\Omega\) |
| External Load | \(R = 12\,\Omega\) | |
2. Circuit Configuration Diagram 🖼️
Since the batteries are in parallel, the standard arrangement is shown below:
+─────[ E1=6V, r1=0.5Ω ]─────+
| |
| +──────[ R = 12Ω ]──────
| |
+─────[ E2=10V, r2=1Ω ]──────+
Parallel connection: Both batteries share the same terminal voltage \(V\).
3. Equivalent Circuit Parameters
A. Equivalent Internal Resistance (\(r_{eq}\))
\[
\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{0.5} + \frac{1}{1} = 2 + 1 = 3\,\Omega^{-1}
\]
\[
\implies r_{eq} = \frac{1}{3}\,\Omega \approx 0.333\,\Omega
\]
B. Equivalent EMF (\(E_{eq}\))
\[
\frac{E_{eq}}{r_{eq}} = \frac{E_1}{r_1} + \frac{E_2}{r_2}
\]
\[
3 E_{eq} = \frac{6}{0.5} + \frac{10}{1} = 12 + 10 = 22
\]
\[
\implies E_{eq} = \frac{22}{3}\,\text{V} \approx 7.333\,\text{V}
\]
4. Total Current (\(I\)) and Terminal Voltage (\(V\))
Total Current (\(I\)):
\[
I = \frac{E_{eq}}{R + r_{eq}} = \frac{\frac{22}{3}}{12 + \frac{1}{3}} = \frac{\frac{22}{3}}{\frac{36}{3} + \frac{1}{3}} = \frac{\frac{22}{3}}{\frac{37}{3}} = \frac{22}{37}\,\text{A}
\]
\[
I \approx 0.595\,\text{A}
\]
Terminal Voltage (\(V\)):
\[
V = I \cdot R = \frac{22}{37} \times 12 = \frac{264}{37}\,\text{V}
\]
\[
V \approx 7.135\,\text{V}
\]
5. Current Through Each Battery (\(I_1\) and \(I_2\))
Using the formula for current in a battery: \(I_{\text{cell}} = \frac{E - V}{r}\)
A. Current Through Battery 1 (\(I_1\)):
\[
I_1 = \frac{E_1 - V}{r_1} = \frac{6 - \frac{264}{37}}{0.5}
\]
\[
= \frac{\frac{222 - 264}{37}}{0.5} = \frac{\frac{-42}{37}}{0.5} = -2 \times \frac{42}{37} = -\frac{84}{37}\,\text{A}
\]
\[
I_1 \approx -2.27\,\text{A}
\]
Note: Negative sign means this battery is being charged.
B. Current Through Battery 2 (\(I_2\)):
\[
I_2 = \frac{E_2 - V}{r_2} = \frac{10 - \frac{264}{37}}{1} = \frac{\frac{370 - 264}{37}}{1} = \frac{106}{37}\,\text{A}
\]
\[
I_2 \approx +2.86\,\text{A}
\]
Note: Positive sign means this battery is supplying current (discharging).
✅ Final Result: Current through the 6V battery is 2.27 A (Charging).
Current through the 10V battery is 2.86 A (Discharging).
Current through the 10V battery is 2.86 A (Discharging).