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Moving Charges And Magnetism

Question
CBSEENPH12039745
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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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