Question

(a) With the help of a diagram, derive the formula for the resultant resistance of three resistors connected in series.
(b) For the circuit shown in the diagram given below:



Calculate:
(i) the value of current through each resistor.
(ii) the total current in the circuit.
(iii) the total effective resistance of the circuit.

Answer 1

(a)

Let the current in the circuit be I amperes and the battery be of strength V volts.
Let the combined resistance of the three resistors be R ohms.
Therefore, according to Ohm's law, we have:
V = IR .... (1)
We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore:
V = V₁ + V₂ + V₃
V = IR₁ + IR₂ + IR₃
V = I (R₁ + R₂ + R₃) ... (2)
From the equations (1) and (2) we have:
R = R₁ + R₂ + R₃

(b) The current through the 5 Ω resistor can be calculated as:
V = IR
I = V/R
I = 6/5 = 1.2 A
The current through the 10 Ω resistor can be calculated as:
I = V/R
I = 6/10 = 0.6 A
The current through the 30 Ω resistor can be calculated as:
I = V/R
I = 6/30 = 0.2 A
(b)The total current in the circuit, I = 1.2 A + 0.6 A + 0.2 A = 2 A

(c) The three resistors are connected in parallel. Therefore:
1/R = 1/R₁ + 1/R₂ + 1/R₃
Here, R₁=5 $\mathrm{\Omega }$,
R₂ = 10 $\mathrm{\Omega }$,
R₃= 30 $\mathrm{\Omega }$
1/R = (1/5) + (1/10) + (1/30)
1/R = (6 + 3 + 1)/30
1/R = 10/30
R = 3 Ω