# PHYSICS HELP PLZ!!!!!!!!!!!!!!?

A 12-V battery is connected to terminals A and B in (Figure 1). Suppose that R = 62 Ω

A) Find the current in 24-Ω resistor.

B)Find the current in 75-Ω resistor.

C)Find the current in resistor R.

D)Suppose the value of R is increased. For each resistor in turn, state whether the current flowing through it increases or decreases.

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Relevance

As I look at the drawing, I can see that the 75 Ω resistor is in parallel with 62 Ω resistor. To determine the equivalent resistance, use the following equation.

1/Req = 1/75 + 1/62

Multiply both sides of this equation by 4,650 * Req.

4,650 = 62 * Req + 75 * Req

137 * Req = 4,650

Req = 4,650 ÷ 137

This is approximately 33.9 Ω. To determine the total resistance of the circuit, add 24 Ω.

Total resistance = 4,650 ÷ 137 + 24 = 4,650 ÷ 137 + 3,288 ÷ 137 = 7,938 ÷ 137

This is approximately 58 Ω.

Let’s use the following equation to determine the current that flows through the 24 Ω resistor.

I = V ÷ R

I = 12 ÷ (7,938 ÷ 137)

I = 1,644 ÷ 7,938

This is approximately 0.21 amp. To determine the voltage for the 24Ω resistor, use the following equation

∆ V = I * R

∆V = (1,644 ÷ 7,938) * 24 = 39,456 ÷ 7,938

The decrease of voltage is approximately 4.97 volts. To determine the voltage for the other two resistors, subtract this number from 12 volts.

V = 12 – 39,456 ÷ 7,938

The voltage for the other two resistors is approximately 7.03 volts. Let’s use this number to determine the current that flows through the other two resistors.

For the 75 Ω resistor, I = (12 – 39,456 ÷ 7,938) ÷ 75

This is approximately 0.094 amp.

For the 62 Ω resistor, I = (12 – 39,456 ÷ 7,938) ÷ 62

This is approximately 0.113 amp.

D)Suppose the value of R is increased. For each resistor in turn, state whether the current flowing through it increases or decreases.

In the parallel circuit, the voltage is the same for both resistors, as the value of R is increased, less current will flow through it To see what happens, let the resistance for R be 80 Ω

1/Req = 1/80 + 1/62

4960 * Req = 62 + 80

Req = 4960 ÷ 142

This is approximately 34.9 Ω.

Total resistance = 24 + 4960 ÷ 142

This is approximately 58.9 Ω. By increasing the total resistance, less current will flow through the 24 Ω resistor. I hope this is helpful for you.

• First get the total equivalent R

75 in parallel with 62 = 62•75 / (62+75) = 33.94 Ω

in series with 24Ω that is 57.94 Ω

total I = 12/57.94 = 0.2071 amps (current in 24Ω)

that produces voltage across parallel pair of 0.2071 x 33.94 = 7.029 volts

which produces current in 75Ω of 7.029 volts/75Ω = 0.0937 amps

which produces current in 62Ω of 7.029 volts/62Ω = 0.1134 amps

increase R, overall current goes down, as total R goes up, which is current in 24Ω. Current in R goes down also. Current in 75Ω may go down initially.