Question

Three resistors of resistances 2 Ω, 3 Ω and 4 Ω are connected in series with a 15 V supply. Calculate the i) total resistance of the circuit i) total current iii) voltage drop across each resistance.​

Answer 1

Given that,

Three resistors of resistances 2 Ω, 3 Ω and 4 Ω are connected in series with a 15 V supply.

Solution,

The equivalent resistance in series combination is given by :

`R=R_1+R_2+R_3\\\\R=2+3+4\\\\=9\ \Omega`

(i) Total resistance of the circuit is equal to 9 ohms.

(ii) Let I be the total current. Using Ohm's law as follows :

`V=IR\\\\I=(V)/(R)\\\\I=(15)/(9)\\\\=1.67\ A`

(iii) Voltage drop across 2 ohm resistor is :

`V=1.67* 2=3.34\ V`

Voltage drop across 3 ohm resistor is :

`V=1.67* 3=5.01\ V`

Voltage drop across 4 ohm resistor is :

`V=1.67* 4=6.68\ V`

Hence, this is the required solution.