Question

Consider the given electric circuit:

Calculate the following:

1. Total resistance of the circuit.
2. The electric current drawn from the battery.
3. Potential difference between points P and Q.

Answer 1

The circuit has a 10 V battery connected between points P and R.

Between P and Q, there are three resistors: 2 Ω, 5 Ω, and 3 Ω.

Between R and S, there are resistors 4 Ω, 5 Ω, and 1 Ω.

The 5 Ω and 3 Ω resistors between P and Q are in series because they are connected end-to-end with no branching.

So, their combined resistance is:

R₅₊₃​ = 5 + 3

= 8 Ω

This 8 Ω resistor is in parallel with the 2 Ω resistor between P and Q.

Calculate the equivalent resistance of 2 Ω and 8 Ω in parallel:

`1/R_(PQ) = 1/2 + 1/8`

= `4/8 + 1/8`

= `5/8`

∴ R_PQ = `8/5` = 1.6 Ω

The 5 Ω and 1 Ω resistors between R and S are in series:

R₅₊₁ ​= 5 + 1

= 6 Ω

This 6 Ω resistor is in parallel with the 4 Ω resistor:

`1/R_(RS) = 1/6 + 1/4`

= `2/12 + 3/12`

= `5/12`

∴ R_RS = `12/5` = 2.4 Ω

a. The resistors between P and Q (1.6 Ω) and between R and S (2.4 Ω) are in series with each other and with the connecting wires.

So, total resistance:

R_total​ = R_PQ​ + R_RS

​= 1.6 + 2.4

= 4.0 Ω

b. Using Ohm’s law:

I = `V/R`

= `10/4`

= 2.5 A

c. Current through the branch P-Q is the same as the total current (since P-Q and R-S are in series).

Voltage drop across P-Q branch:

V_PQ​ = I × R_PQ​

= 2.5 × 1.6

= 4.0 V