Question

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.12 A. When the resistors are connected in parallel to the battery, the total current from the battery is 9.39 A. Determine the two resistances.

Answer 1

Answer:

R = 9.85 ohm , r = 0.85 ohm

Explanation:

Let the two resistances by r and R.

when they are connected in series:

V = 12 V

i = 1.12 A

The equivalent resistance when they are connected in series is

Rs = r + R

So, By using Ohm's law

V = i Rs

Rs = V / i = 12 / 1.12 = 10.7 ohm

R + r = 10.7 ohm    .... (1)

When they are connected in parallel:

V = 12 V

i = 9.39 A

The equivalent resistance when they are connected in parallel

$R_{p}=\frac{R+r}{rR}$

So, By using Ohm's law

V = i Rp

Rp = V / i = 12 / 9.39 = 1.28 ohm

$\frac{R+r}{rR}=1.28$    .... (2)

by substituting the value of R + r from equation (1) in equation (2), we get

r R = 8.36 ..... (3)

$R-r = \sqrt{\left ( R+r \right )^{2}-4rR}$

$R-r = \sqrt{\left ( 10.7 \right )^{2}-4\times 8.36}=9$ ..... (4)

By solvng equation (1) and (4), we get

R = 9.85 ohm , r = 0.85 ohm