Question

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Q) A cell of emf 2v and interanal resistance 1.2 ohm and is connected to an ammeter of resistance 0.8 ohm and two resistors of 4.5 ohm and 9 ohm as shown in figure below

Find the potential difference accross the 4.5 ohm resistor .​

Answer 1

Answer:

Given,

Resistance = 1.2, 4.5, 0.8, 9ohms

Voltage = 2V

We know that 4.5ohm and 9ohm resistors are in parallel

Req = 1/R1 + 1/R2

= 1/4.5+1/9

= 3ohm

0.8ohm and 3ohm are in series

Req = R1+R2

= 0.8+3

= 3.8ohm

1.2ohm and 3.8ohm are connected in series

= 1.2+3.8

= 5ohm

We know that V = IR

I = V/R

= 2/5

= 0.4amp.

Potential difference V = IR

= 0.4×4.5

= 1.8V.

Potential difference across 4.5 resistor is 1.8V.