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Please solve this question 28. The circuit diagram in Fig. 8.53 s ows three resistors 2 Q, 4 Q, and R Q connected to a battery of e.m.f. 2 V and internal resistance 3 Q. If main current of 0•25 A flows through the circuit, find : (a) the p.d. across the 4 Q resistor, (b) across the internal resistance of the cell, (c) the p.d. across the R Q or 2 Q resistor, and (d) the value of R. 025 A Fig. 8.53 Ans. (a) 1 v (b) 0-75 V (c) 9.25 V (d) 2 Q form chem frorr use as ch

Hello Prachi dear, R value has to be 2 ohm
How?
As 0.25 A current flows in the circuit with emf 2 V then net resistance in the circuit becomes as 2/0.25 = 8 ohm
Already 3 ohm as internal, 4 ohm in series so remaining has to be 1 ohm
This 1 ohm has to be with 2 ohms (two in number) in parallel
Hence R = 2 ohm 
P.D across 3 ohm will be 3 * 0.25 = 0.75 V
Hence balance 1.25 V will be in total across parallel combination and across 4 ohm
So p.d across 4 ohm will be 1 V and balance 0.25 V will be across parallel combination
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a.R=4ohm
I=0.25A
V=IR=(4?0.25)V=1V ans
b.V=IR=0.25?3=0.75V ans
c.Effectiveness of resistance ofa parallel combination of 2 ohm=1ohm
V=I/R=0.25/1=0.25V ans
d. I=0.25A
e.m.f.=I(R'+r)
=>2=0.25(R'+3)
=>R'=5W
=>(2R/2+R)+4=5
=>R=2ohm ans
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