A p.d. of 6 V is applied to two resistors of 3 Ω and 6 Ω connected in parallel. Calculate:

(a) the combined resistance
(b) the current flowing in the main circuit
(c) the current flowing in the 3 Ω resistor.

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Solution

(a) The resistors of 3 Ω and 6 Ω are connected in parallel. Therefore, their net resistance can be calculated as:
1/R = 1/R₁ + 1/R₂
Here, R₁ = 3 Ώ
R₂ = 6 Ώ
So:
1/R = (1/3) + (1/6)
1/R = (2 + 1)/6
1/R = 3/6
R = 2 Ω

(b) The current flowing through the main circuit,
I = V/R
I = 6/2 A
I = 3 A

(c) The current flowing in the 3 Ω resistor,
I = V/R
I = 6/3 = 2 A

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Q. A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω connected in series. Calculate:

(a) the combined resistance
(b) the current flowing
(c) the p.d. across the 6 Ω resistor

Q. The p.d. across a 3 Ω resistor is 6 V. The current flowing in the resistor will be:

(a)

\frac{1}{2} A

(b) 1 A
(c) 2 A
(d) 6 A

A battery of e.m.f. 15 V and internal resistance $3 Ω$ is connected to two resistors $3 Ω$ and $6 Ω$ connected in parallel. Find:

(a) the current through the battery,

(b) the p.d. between the terminals of the battery,

(d) the current in $6 Ω$ resistor.

In the circuit given below:

(a) What is the combined resistance?
(b) What is the p.d. across the combined resistance?
(c) What is the p.d. across the 3 Ω resistor?
(d) What is the current in the 3 Ω resistor?
(e) What is the current in the 6 Ω resistor?