In the circuit shown, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 V.

Options

A. The resistor `R_5` that dissipates maximum power.
B. The resistor `R_1` that dissipates maximum power.
C. The potential difference across resistor `R_3 is 0.4V.`
D. The current passing through 3 V battery is 40 mA.

Answer

B, C, D

Explanation

b., c., d.
The equivalent circuit can
be redrawn as shown in figure.
From the figure it is
obvious that power dissipated
by `R_1` is maximum. Potential
difference across `R_2 is 25/(25+50) xx 3V = 1V`.
Therefore, potential difference across `R_2 and R_4 is 20/(20+30) xx 1V`
`= 0.4V.`
The equivalent resistance of circuit across the cell is `50 + 25 =`
`75 Omega`
Therefore, current drawn through cell is
`3/75 xx 1000mA = 40mA`.