fundamental of electrical circuits

Question

Answer 1

The ratio  of their resistance is equal

\(\frac{P}{Q}=\frac{R}{S}\)

\(\frac{30}{15}=\frac{10}{5}\)

2 = 2

Wheatstone condition

Then PQ are in series

R = 30 + 15

R = 45 Ω

and

RS are also series

R = 10 + 5

R = 15 Ω

Then,

\(\frac{1}{R}\) = \(\frac{1}{45}\) + \(\frac{1}{15}\)

\(\frac{1}{R}\) = \(\frac{1\ +\ 3}{45}\)

R = \(\frac{45}{4}\)Ω

R_T = 5 + \(\frac{45}{4}\)

R_T = 16.25 Ω