Question

Three resistors R1 = 5 Ω, R2 = 8 Ω and R3 = 12 Ω are connected in
series with a battery of 4 V. Find the value of current in the circuit
and the voltage drop across each resistor.

Answer 1

Answer:

0.16

Explanation:

Rs=R1+R2+R3

=5+8+12

=25

I=V÷Rs

=4÷25

=0.16

Answer 2

Solution :

Given :

Three resistances of 5 Ω, 10 Ω and 30 Ω are connected in parallel with a battery of 12 V.

To Find :

• Current in circuit .
• The voltage drop across each resistor.

Step by step explanation

1) We have to find Current in circuit.

First we have to find Equivalent resistance of the circuit .

$\blue{R_{eq}=R_1+R_2+R_3}$

Now put the values then ,

Equivalent resistance 25 Ω

Now, E= 4

R = 25 Ω

Now , by ohms law

E= IR

I = 0.16

Hence , 0.16 A current will flow through the circuit.

2) We have to find voltage drops each resistor.

Since all three resistances are connected in series, so same current will flow through each resistor .

Voltage drop across 5 Ω

E = IR

= 0.16× 5 = 0.8 V

Voltage drop across 8 Ω

E = IR

= 0.16× 8 = 1.28 V

Voltage drop across 12Ω

E = IR

= 0.16× 12= 1.92 V