Question

find the total reristance in the circuit​

Answer 1

12 ohm is the required answer.

GIVEN:

• R1 = 8 ohm
• R2 = 12 ohm
• R3 = 7.2 ohm

TO FIND:

Total resistance in the circuit, R

FORMULA:

• When the resistors are connected in series, the effective resistance is given by, $R = R_1 + R_2 +R_3+...$

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

SOLUTION:

STEP 1:

R1 and R2 are connected in parallel . Let their effective resistance be Rs.

$\dfrac{1}{R_s} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \\ \\ \dfrac{1}{R_s} = \dfrac{1}{8} + \dfrac{1}{12} \\ \\ \dfrac{1}{R_s} = \dfrac{12 + 8}{12 \times 8} \\ \\ \dfrac{1}{R_s} = \dfrac{20}{12 \times 8} \\ \\ R_s = \dfrac{12 \times 8}{20} \\ \\ R_s = \dfrac{12}{20} \times 8 \\ \\ R_s = \dfrac{3}{5} \times 8 \\ \\ R_s = \dfrac{3 \times 8}{5} \\ \\ R_s = \dfrac{24}{5} \\ \\ R_s = 4.8 \: ohm$

STEP 2

Rs and R3 are in series. Let their effective resistance be R.

$R = R_s + R_3 \\ \\ R = 4.8 + 7.2 \\ \\ R = 12 \: ohm$

ANSWER:

The total effective resistance of the circuit is 12 ohm.

SOME POINTS TO NOTICE:

• When the resistors are connected in series, their effective resistance is higher than individual resistance.

• When the resistors are connected in parallel, their effective resistance is lower than individual resistance.