The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in series with R, the current drops to 4.72 A. What is R?

Answer: R is equal to 10.12Ω.

Explanation: From the question, the second single loop circuit has 2 resistance in series. One is known R¹ while the other is unknown R. For resistance in series, the individual resistance given will be added together to obtain the overall resistance. The voltage of both single loop circuit are constant. Using the ohnm's law:

V = IR

The first circuit, V = I¹R

The 2nd circuit, V = I² (R + R¹)

V = is constant for the both circuit. Therefore,

I¹R=I² (R+R¹).

Making R the subject of formula,

R = I²*R¹ / I¹ - I ²

Where, I² = 4.72A, R¹=3.4Ω, I¹ = 6.3, I²=4.72A.

Therefore R = 4.72 * 3.4/6.3-4.72

R = 16.048/1.58

R = 10.12Ω.