Question

Solve this:

156. Three resistors having resistances of 1.6 $\Omega$, 2.4 $\Omega$ and 4.8 $\Omega$ are connected in series to a 22 V battery (ideal). Find

(a) the current in each resistor
(b) the potential difference across each resistor
(c) the total current through the battery.
(d) the power dissipated by each resistor
(e) Which resistor dissipates maximum power

Ans- (a) 205 A (b) 4 V, 6 V, 12 V (c) 2.5 A (d) 10 wW, 15W, 30 W
(e) 4.8 $\Omega$

Answer 1

Dear student,
The three resistors are connected in series, so,
Req=R1+R2+R3
Req=1.6+2.4+4.8
Req=8.8ohms
Given the value of the voltage of the battery is 22V.

(a)Since all the resistors are connected in series, the current will also remain same in each resistor. Hence,
I=V/R
I=22/8.8
I=2.5A

(b)The potential difference across each resistor will be
1)at 1.6 ohm
V=IR
V=2.5 x 1.6
V=4V

2)at 2.4ohm
V=IR
V=2.5x2.4
V=6V

3)at 4.8ohm
V=2.5x 4.8
V=12V

(c)The total current through the battery will be
I=V/R
I=22/8.8
I=2.5A

(d)The power dissipated by each resistor be calculated by a formula
P=I²R
1)for 1.6 ohm
P=(2.5)²x 1.6
P=10W

2) for 2.4 ohm
P=(2.5)²x 2.4
P=15W

3) for 4.8 ohm
P=(2.5)²x 4.8
P=30W

(e)From above values of the power, it can be seen that the resistor of 4.8 ohms dissipates maximum power.
Regards