Question

For the circuit shown in the figure
Calculate
(i) the total effective resistance of the circuit,
(ii) the total current drawn from the battery​

Answer 1

GIVEN:

• R1 = 12 ohm
• R2 = 18 ohm
• R3 = 30 ohm
• Potential difference, v = 6 volt

TO FIND

• Total effective resistance, R
• Total current drawn from the battery, I

FORMULAE:

• According to ohm's law, V = IR

• When resistors are connected in parallel, effective resistance is given by $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

SOLUTIONS:

STEP 1: To find effective resistance

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \\ \\ \frac{1}{R} = \frac{1}{12} + \frac{1}{18} + \frac{1}{30 } \\ \\ LCM \: of \: 12 ,\: 18 ,\: 30 \: is \: 180 \\ \\ \dfrac{1}{R} = (\dfrac{1 \times 15}{12 \times 15} ) + ( \dfrac{1 \times 10}{18 \times 10} ) + ( \dfrac{1 \times 6}{30 \times 6} ) \\ \\ \dfrac{1}{R} = \dfrac{15}{180} + \dfrac{10}{180} + \dfrac{6}{180} \\ \\ \dfrac{1}{R} = \dfrac{15 + 10 + 6}{180} \\ \\ \frac{1}{R} = \frac{31}{180} \\ \\ R = \frac{180}{31} \\ \\ R = 5.81 \: ohm$

Total effective resistance is 5.81 ohm.

STEP 2: To find total current

$V = IR \\ \\ 6 = I \times 5.81 \\ \\ I = \frac{6}{5.81} \\ \\ I = 1.03 \: ampere$

Total current drawn from the battery is 1.03 A.

ANSWER:

• Total effective resistance is 5.81 ohm.

• Total current drawn from the battery is 1.03 A.

IMPORTANT POINTS:

• When the resistors are connected in series, the current is same and voltage is different.

• When the resistors are parallel, the potential difference is same but current is different across individual resistors.

Answer 2

Answer:

(a) 180/31 or 5.806 ohm

(b) 1.03 A

Explanation: Calculate equivalent resistance,as all resistances are in parallel combination. Then use ohm's law(V=IR) to calculate the current of close circuit.