Question

The ratio of the currents i$_1$, i$_2$ and i$_3$ in the circuit is:

• A. 1 : 1 : 1
• B. 3 : 2 : 1
• C. 1 : 2 : 3
• D. 3 : 1 : 2

Hint:

Use Kirchhoff’s laws for circuit problems: KCL for current at junctions, KVL for voltage loops. Solve for currents and check ratios.

Answer 1

The Correct Option is B

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.