Answer :

Given:

Effective resistance when two resistors are connected in series =80 Ω

Let us consider the resistances as R_{1} and R_{2 }in series combination,

V = V_{1 }+ V_{2 }and V=IR

As we know that in series connection current is constant, we get

R= R_{1} + R_{2} = 80 Ω ---(1)

Effective resistance when two resistors are connected in parallel combination = 20Ω

We get

........(2)

putting the value of equation(1) in equation(2)

R_{1}R_{2} = 1600

(R_{1}-R_{2})^{2} = (R_{1} + R_{2})^{2} - 4R_{1}R_{2}

= 80^{2} - (4×1600)

= 0

R_{1}-R_{2} = 0 ---(3)

adding (1) and (3)

R_{1=}40

Subtracting (1) and(3)

R_{2}=40Ω

we get R_{1} = R_{2} = 40Ω

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