In the circuit shown, the 15- resistor drops a voltage of 4.0 V. (A) Find the ba
ID: 1366935 • 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
a)
current through 15 ohms = 4/15
= 0.267 A
current through 30 ohms = 4/30
= 0.133 A
current through 40 ohms = 0.267 + 0.133
= 0.4
so, battery voltage = I*(40+10)
= 0.4*(40 + 10)
= 20 volts <<<<=---------Answer
B) Power dissipated by 40 ohm = I^2*R
= 0.4^2*40
= 6.4 Watts <<<<=---------Answer
C) Current through 60 ohms = 20/60
= 0.333 A
so, total current = 0.4 + 0.333
= 0.7333 A <<<<=---------Answer