Question

In the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R₁? 

Answer 1

5 ohm, 10 ohm and R₁ are in series

`1/"R"_"p" = 1/5 + 1/10 + 1/"R"_1`

`1/"R"_"p" = (2+1)/10 + 1/"R"_1`

`= 3/10 + 1/"R"_1`

`1/"R"_"p" = (3 "R"_1 + 10)/10 "R"_1`

`"R"_"p" = 10 "R"_1/(3 "R"_1 + 10)`

Now, 6 ohm, 6 ohm and R_p are in series

Thus,

`"R"_"eq" = 12 + 10 "R"_1/(3 "R"_1 + 10)`    ....(1) 

V = I R_eq

From the circuit

R_eq = `30/2` = 15 A   ...(2)

Equating (1) and (2)

`therefore 12 + 10 "R"_1/(3 "R"_1 + 10)` = 15

`therefore 10 "R"_1/(3"R"_1 +10)` = 3

`therefore 10 "R"_1 = (9 "R"_1 + 30)`

Thus, R₁ = 30 ohm.