Question

A cell of e.m.f. 1.5V and internal resistance 1.0Ω is connected to two resistors of 4.0Ω and 20.0Ω in series as shown in the figure:

1. current in the circuit. 

2. potential difference across the 4.0ohm resistor. 

3. voltage drop when the current is flowing. 

4. potential difference across the cell.

Answer 1

Here e.m.f., E = 1.5 V 

Internal resistance 1.0Ω 

All the resistances are connected in series

The total circuit resistance, R= 1 + 4 + 20 = 25Ω

The current,i = E/R =1.5/25 = 0.06A 

Potential difference across 4Ω resistance = r × i = 4 × 0.06 = 0.24V 

Voltage drop across the cell = 0.06 × 1 = 0.06V

Potential difference across the cell = 1.5 — 0.06 = 1.44V