A 30.0 Ohm resistor and a 40.0 Ohm are connected in series to a 9.00 V power sup

Question

A 30.0 Ohm resistor and a 40.0 Ohm are connected in series to a 9.00 V power supply. (a) What is the total resistance of the combination? Show your calculations here and record your answer as calculated R_eq at the top of table 2. (e) What are the voltage drops across R_1, R_2, and R_eq? Show your calculations here and record your answers as calculated V_1, V_2 and V_eg in table 3. (f) What are the currents though R_1, R_2, and R_eg? Show your calculations here and record your answers as calculated I_1, I_2 and I_eq in table 3.

Explanation / Answer

Given R1 = 30 ohm

R2 = 40 ohm

v = 9 V

a)

Req = R1 + R2 = 30 + 40

Req = 70 ohm

e)

total current I = v / Req

I = 9 / 70

I = 0.128 A

V1 = I * R1 = 0.128 * 30

V1 = 3.84 V

V2 = I * R2 = 0.128 * 40

V2 = 5.12 V

Veq = 9 V

f)

R1 and R2 are in series the total current is same across R1 and R2

I1 = I2 = Ieq = 0.128 A

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