Question

Helloo. I need help with this Physics question:

Three resistors of resistance 4 ohm, 6 ohm and 12 ohm are connected in parallel. This parallel arrangement is connected in series with a resistance of 1 ohm and a cell of e.m.f 1.5V. Calculate the current in the circuit.

Answer 1

0.5 A.

Step-by-step explanation:

The three resistors of resistance 4 ohm, 6 ohm and 12 ohm connected in parallel will have an equivalent resistance of:

`\tt (1)/(R)=(1)/(4)+(1)/(6)+(1)/(12)`

`\tt (1)/(R)=(1*6+1*4+1*2)/(24)=(12)/(24)=(1)/(2)`

while doing Criss cross multiplication:

we get,

R=2 ohm

The equivalent resistance of the parallel arrangement is connected in series with a resistance of 1 ohm,

so the total resistance of the circuit is 2+1= ohm.

The current in the circuit is calculated using the following formula:

`\boxed{\boxed{\tt I =(V)/(R)}}`

Where:

• I is the current
• V is the voltage
• R is the resistance

In this case, the voltage or e.m.f is 1.5 V and the resistance is 2 ohms, so the current is:

`\tt I =( 1.5 \:V )/(3\: ohm) = 0.5 A`

Therefore, the current in the circuit is 0.5 A.