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

a. In the upper branch, the 4 Ω and 1 Ω resistors are in series.

⇒ 4 + 1 = 5Ω

This 5 Ω branch is in parallel with the other 5 Ω resistor.

The equivalent resistance between R and S:

R_RS = `(5 xx 5)/(5 + 5)`

= `25/10`

= 2.5 Ω

In the lower branch, 3 Ω and 2 Ω resistors are in series.

⇒ 3 + 2 = 5 Ω

This 5 Ω branch is in parallel with the middle 5 Ω resistor.

Equivalent resistance between P and Q:

R_PQ = `(5 xx 5)/(5 + 5)`

= `25/10`

= 2.5 Ω

The two networks are connected in series.

∴ Total resistance (R_total) = R_RS + R_PQ

= 2.5 + 2.5

= 5 Ω

b. By using Ohm’s law:

V = IR

I = `V/R_"total"`

= `10/5`

= 2 A

c. The total current of 2 A flows through the equivalent resistance of the P-Q section (R_PQ).

V_PQ = I × R_PQ

= 2 × 2.5

= 5 V