An electrical circuit is given below. V _1=1 ~V and Resistance R =1000 . The current through the resistance R is very close to 1 mA and the voltage across point A and B , V _ AB =1 ~V . Now the circuit is changed to: where value of V _2=5 ~V . The internal resistances of both the batteries are 0.1 . The current through the resistance R is about:
C. 3.0 mA
aligned & Equivalent voltage = _1 r_1+ _2 r_2 r_1+r_2 \\ & aligned & = 1 0.1+5 0.1 0.1+0.1 \\ E _ eq & = 0.6 0.2 =3 ~V aligned aligned From first circuit, we have i= R +r Value of r is very less can be neglected as compared to R Now, aligned i & = R \\ 10^ -3 & = 1 R \\ R & =10^3=1000 aligned Now, from second circuit aligned i_1 & = E _ eq R +r_ e q = 3 1000 \\ i_1 & =3 10^ -3 \\ & =3 ~mA aligned
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