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Question

Physics Question on Current electricity

In the circuit R1=R2R_1 = R_2.The value of EE and R1R_1 are ________(EEMF,R1E-EMF,R_1-resistance)

A

180V,60Ω180\, V, 60\, \Omega

B

120V,60Ω120\, V,60\, \Omega

C

180V,10Ω180\, V,10\, \Omega

D

120V,10Ω 120\,V, 10\,\Omega

Answer

180V,60Ω180\, V, 60\, \Omega

Explanation

Solution

Applying Kirchoff's 2 nd law in loop ABEFAA B E F A
IR(I1.5)R+E=0-I R-(I-1.5) R+E=0

or E=IR+(I1.5)R E=I R+(I-1.5) R
E=R[2/1.5]...(i)\Rightarrow E=R[2 /-1.5] \,\,\,\,\,\,...(i)
Now apply KVL in loop ABCDEFAABCDEFA, IR1.5R+E=0-I R-1.5 R'+E=0
E=IR+1.5R...(ii)\Rightarrow E=I R+1.5 R'\,\,\,\,\,\,\,...(ii)
Eliminating / from Equ. (i) and (ii),
12[ER+1.5]=1\Rightarrow \frac{1}{2}\left[\frac{E}{R}+1.5\right]=1\,\,\,\,\,[from E(i)]
and [E1.5RR]=1\left[\frac{E-1.5 R^{\prime}}{R}\right]=1
[from E (ii)]
So, ER+1.5=2R[E1.5R]\frac{E}{R}+1.5=\frac{2}{R}\left[E-1.5 R'\right]
So, if RR' were known, then from options we could get the answer.