The ratio of the currents i$_1$, i$_2$ and i$_3$ in the circuit is:
The ratio of the currents i$_1$, i$_2$ and i$_3$ in the circuit is:
Show Hint
- 1 : 1 : 1
- 3 : 2 : 1
- 1 : 2 : 3
- 3 : 1 : 2
The Correct Option is B
Solution and Explanation
The circuit has a 70 V battery, a 20 $\Omega$ resistor (i$_1$), splitting into two branches: 20 $\Omega$ (i$_2$) to B (0 V) and 30 $\Omega$ (i$_3$) to C (10 V).
Apply Kirchhoff’s laws: At junction D, $i_1 = i_2 + i_3$.
Voltage drop across 20 $\Omega$ (i$_1$): $70 - i_1 \cdot 20 = V_D$.
Branch DB: $V_D - i_2 \cdot 20 = 0 \implies V_D = 20 i_2$.
Branch DC: $V_D - i_3 \cdot 30 = 10 \implies V_D = 30 i_3 + 10$.
Equate: $20 i_2 = 30 i_3 + 10 \implies 2 i_2 = 3 i_3 + 1$.
From $i_1 = i_2 + i_3$, and solving: $2 i_2 = 3 i_3 + 1 \implies i_2 = \frac{3 i_3 + 1}{2}$.
Test ratio 3:2:1: Let $i_3 = 1$, then $i_2 = \frac{3 \cdot 1 + 1}{2} = 2$, and $i_1 = 2 + 1 = 3$.
Ratio $i_1 : i_2 : i_3 = 3 : 2 : 1$.
Verify: $V_D = 20 i_2 = 20 \cdot 2 = 40$, and $V_D = 30 i_3 + 10 = 30 \cdot 1 + 10 = 40$, which matches.
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