Question

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 1

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I'  and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

Answer 2

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.