Question

A network of resistors is connected to a 14 V battery with internal resistance 1 Q as shown in the circuit diagram.

i. Calculate the equivalent resistance,

ii. Current in each resistor,

iii. Voltage drops V_AB, V_BC and V_DC.

Answer 1

For equivalent resistance (R_eq):

R_AB is given as,

∴ R_AB = 2 Ω

R_BC = R₃ = 1 Ω

Also, R_CD is given as,

∴ R_CD = 3 Ω

∴ R_eq = R_AB + R_BC + R_CD

= 2 + 1 + 3 = 6Ω

ii. Current through each resistor:

Total current, I = \(\frac{E}{R_{eq}+r}=\frac{14}{6+1}\) = 2 A

Across AB, as, R₁ = R₂ 

V₁ = V₂ 

∴ I₁ × 4 = I₂ × 4 

∴ I₁ = I₂

But, I₁ + I = I 

∴ 2I₁ = I 

∴ I₁ = I₂ =1 A ….(∵I = 2 A) 

Similarly, as R₄ = R₅ 

I₃ = I₄ = 1 A 

Current through resistor BC is same as I. 

∴ I_BC = 2 A 

iii. Voltage drops across AB, BC and CD: 

V_AB = IR_AB = 2 × 2 = 4 V

V_BC = IR_BC = 2 × 1 = 2 V

V_CD = IR_CD = 2 × 3 = 6 V