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

The Total Effective Resistance is $3\ \Omega$

Explanation:

Let the current 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 -----(i)

We know that when the resistors are connected in series, the current is the same in all the resistors. Therefore

$V = V_1 +V_2+ V_3$ ------------------ (ii)

Let the current flowing through the whole circuit is I, and the equivalent resistance be R.

According to ohm's law

V = IR

$V_1 = IR_1$ -------- (iii)

$V_2 = IR_2$-------- (iv)

$V_3 = IR_3$ ------ (v)

By using eq (i),(ii),(iii) ,(iv) and (v) we get

$IR= IR_1+ IR_2+ IR_3$

$R= R_1+ R_2+ R_3$

(b)(i) Here $R_1=5\ \Omega$, $R_2= 10\ \Omega$ and $R_3= 30\ \Omega$

The value of current through each resistor is different as they are arranged in a parallel connection.

So the current through 5 ohm resistor, $I_1 = \frac {V}{R} = \frac {6}{5}$ = 1.2A

So the current through 10 ohm resistor, $I_2 = \frac {V}{R} = \frac {6}{10}$ = 0.6A

So the current through 30 ohm resistor, $I_3 = \frac {V}{R} = \frac {6}{30}$ = 0.2A

• (ii) The total current in the circuit = $I= I_1+ I_2+ I_3$

$1.2A + 0.6A + 0.2A = 2A$

(iii) The total resistance in the circuit = R, As the circuit is in parallel connection, so

$R= 3\ \Omega$