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Question
In the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R1?
Solution
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 Rp are in series
Thus,
`"R"_"eq" = 12 + 10 "R"_1/(3 "R"_1 + 10)` ....(1)
V = I Req
From the circuit
Req = `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.
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