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In the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R1? |
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Answer» 5 ohm, 10 ohm and R1 are in series 1/Rp = 1/5+1/10+1/R1 1/Rp = (2+1)/10 + 1/ R1 = 3/10 +1/ R1 1/Rp = (3 R1 +10)/10 R1 Rp= 10 R1 / (3 R1 +10) Now, 6 ohm, 6 ohm and Rp are in series Thus, Req = 12 + 10 R1 / (3 R1 +10) ------------------ (1) V = I Req From the circuit Req = 30/2 = 15 A ---------------(2) Equating (1) and (2) 12 + 10 R1 / (3 R1 +10) = 15 10 R1 / (3 R1 +10) = 3 10 R1= (9 R1 + 30) Thus, R1 = 30 ohm. |
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