Question

(a) With the help of a circuit diagram, deduce the equivalent resistance of two resistances connected in series.
(b) Two resistances are connected in series as shown in the diagram:



(i) What is the current through the 5 ohm resistance?
(ii) What is the current through R?
(iii) What is the value of R?
(iv) What is the value of V?

Answer 1

(a)

(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 resistors are connected in series, the current is the same in all the resistors but the voltage is different across each resistor.
Therefore:
V = V₁ + V₂
V = IR₁ + IR₂
V = I (R₁ + R₂) ... (2)
From the equations (1) and (2) we have:
R = R₁ + R₂

(b)
(i) The current through the 5 Ω resistance can be obtained using the equation,
I = V/R
or I = 10/5 = 2 A
(ii) In a series arrangement, the current remains the same across each resistance. Therefore, the current through R also will be 2 A.

(iii) Value of R = V/I
R = 6/2 = 3 Ω
(iv) The value of V = IR
V = 2 x (3 + 5) = 16 V