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Moving Charges And Magnetism

Question
CBSEENPH12039745
Wired Faculty App

Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R1 and R2 (R2>R1). If the potential difference across the source having internal resistance R2 is zero, then

  • straight R space equals space fraction numerator straight R subscript 2 space straight x space left parenthesis straight R subscript 1 space plus straight R subscript 2 right parenthesis over denominator left parenthesis straight R subscript 2 minus straight R subscript 1 right parenthesis end fraction

  • R = R2-R1

  • straight R space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis end fraction
  • straight R space equals space fraction numerator straight R subscript 1 straight R subscript 2 over denominator left parenthesis straight R subscript 1 minus straight R subscript 2 right parenthesis end fraction

Solution

B.

R = R2-R1

The equivalent resistance of the  circuit
Req = R1  + R2 + R
I = 2E/ R1 +R2 +R
according to the questions
-(VB-VA) = E- IR2
0 = E- IR2E= IR2
E = 2ER2/R1+R2+R
R1+R2+R =2R2
R= R1-R2

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω   N1 = 30,
A1 = 3.6 x 10–3 m2,  B1 = 0.25 T
R2 = 14 Ω;  N2 = 42,
A2 = 1.8 x 10–3 m2, B2 = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

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