Question

A cell of emf \( 2 V \) and internal resistance \( 3 \Omega \) is connected in series with two resistors of resistances \( 20 \Omega \) and \( 17 \Omega \) respectively. Determine 

(i) the current in the circuit 

(ii) potential difference across the \( 17 \Omega \) resistor 

(iii) the terminal potential difference of the cell. 

[Ans. (i) \( 0.05 A \) (ii) \( 0.85 V \) (iii) \( 1.85 V \) ]

Answer 1

Determine the Total Resistance in the Circuit

The circuit consists of:

• A cell with an emf (E) of 2V and internal resistance (r) of 3 Ω.
• Two resistors (R₁​ and R₂​) with resistances 20 Ω and 17 Ω respectively, connected in series.

Since the resistors are in series, the total external resistance (R_total) is the sum of the individual resistances:

R_total​ = R₁ ​+ R_2 ​

= 20Ω + 17Ω 

= 37Ω

Calculate the Total Resistance Including the Internal Resistance of the Cell

The total resistance in the circuit (R_eq) is the sum of the internal resistance and the external resistance:

R_eq = r + R_total 

= 3 Ω + 37 Ω

= 40 Ω

Determine the Current in the Circuit

Using Ohm's Law, the current (I) in the circuit can be calculated as:

I = E/Req

= 2 V/40 Ω 

= 0.05 A

Determine the Potential Difference Across the 17 Ω Resistor

The potential difference (V₁₇) across the 17 Ω resistor can be calculated using Ohm's Law again:

V₁₇ = I × R₂

= 0.05 A × 17 Ω

= 0.85 V

Determine the Terminal Potential Difference of the Cell

The terminal potential difference (V_terminal) of the cell is the emf of the cell minus the voltage drop across the internal resistance:

V_terminal = E − I × r

= 2 V − (0.05 A × 3 Ω)

= 2 V − 0.15 V

= 1.85 V

(i) Current in the circuit: I = 0.05 A

(ii) Potential difference across the 17 Ω resistor: V₁₇ = 0.85 V

(iii) Terminal potential difference of the cell: V_terminal = 1.85 V