In the circuit shown, the 15- resistor drops a voltage of 4.0 V. (A) Find the ba
ID: 1366909 • Letter: I
Question
In the circuit shown, the 15- resistor drops a voltage of 4.0 V. (A) Find the battery voltage. You should not have to calculate any currents. ___V. (B) Determine the power dissipated by the 40- resistor. ___W. (C) Find the total current supplied by the battery. Keep two significant figures. ____A
Hint: the 30- and 15- resistors may be combined to an equivalent 10- resistor. Since this is in series with the 40-&Omega resistor, the 40- resistor must have 4 times the voltage drop of the 10-resistor. The sum of these must equal the battery voltage.
Explanation / Answer
given,
resistance = 15 ohm has a voltage drop of 4 V
15 and 30 ohm resistance are parallel so
1 / R1 = 1 / 15 + 1 / 30
R1 = 10 ohm
40 ohm and R1 are in series so
R2 = 40 + 10
R2 = 50 ohm
R2 and 60 ohm resistance are in parallel so,
1 / Rnet = 1 / 60 + 1 / 50
Rnet = 27.27 ohm
current through 15 ohm resistance = 4 / 15
current through 15 ohm resistance = 0.267 A
0.267 = I1 * 15 / (30 + 15)
I1 = 0.801 A
0.801 = I * 50 / (50 + 60)
I = 1.7622 A
V = IR
V = 1.7622 * 27.27
V = 48.055 Volts
battery voltage = 48.055 Volts
power dessipated by 40 ohm resistance = I^2 * R
power dessipated by 40 ohm resistance = 0.801^2 * 40
power dessipated by 40 ohm resistance = 25.66 W
total current supplied by the battery = 1.7622 A