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:

Step-by-step solution

aligned & Equivalent voltage = ₁ r₁+ ₂ r₂ r₁+r₂ & 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 i₁ & = E _ eq R +r_ e q = 3 1000 i₁ & =3 10⁻³ & =3 ~mA aligned