Question

Three resistors of value 4 ω, 10 ω and 7 ω are placed in parallel. they are connected to a 6 ω resistor in series. what is the equivalent resistance in the circuit?

Answer 1

resistance of 4 ohm, 10 ohm and 7 ohm is connected in parallel.

so net resistance in parallel is given by

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

$\frac{1}{R} = \frac{1}{4} + \frac{1}{10} + \frac{1}{7}$

$R = \frac{140}{69} ohm$

now it is connected to 6 ohm resistance in series

so net resistance is given by

$R_{net} = R_1 + R_2$

$R_{net} = 6 ohm + \frac{140}{69} ohm$

$R_{net} = 8.03 ohm$