Question

Two sources of equal emf are connected in series. This combination is, in turn connected to an external resistance R. The internal resistance of two sources are r₁ and r₂ (r₂  > r₁). If the potential difference across the source of internal resistance r₂ is zero, then R equals to:

विकल्प

• `(r_1 + r_2)/(r_2 - r_1)`

• r₂ − r₁

• `(r_1r_2)/(r_2 - r_1)`

• `(r_1 + r_2)/(r_1r_2)`

Answer 1

r₂ − r₁

Explanation:

Net emf = E_net

= E + E

= 2E

Net resistance = R_net

= R + r₁ + r₂

Current in circuit = I 

= `("E"_"net")/("R"_"net")`

= `(2E)/((R + r_1 + r_2))`

Since the potential difference of the cell of internal resistance r₂ is 0,

E − ir₂ = 0

or, `(E - 2ER_2)/(R + R_1 + r_2) = 0`

or, `E = (2Er_2)/((R + r_1 + r_2))`

or, `1 = (2r_2)/((R + r_1 + r_2))`

∴ R = r₂ − r₁