5. All the resistors in this circuit, the 18 Ω resistor included, are in parallel. The voltage drop across each resistor is the same, which is the equivalent to the potential difference of the battery. That is, 36 V [choice K].
6. The ammeter is in series with the branch containing a 12 Ω resistor. Since the voltage drop across this resistor is 36 V, the current in this branch will be I = V/R = (36 V)/(12 Ω) = 3.0 A [choice D].
7. The equivalent resistance of the circuit can be calculated as follows:
1/R = 1/12 + 1/9 + 1/18 = 1/4; thus, R = 4.0 Ω [choice G].
8. Electrical power can be defined as electrical work/time, and electrical work can be given by W = qV. If P = qV/t and q/t = I (charge flowing per unit time is equal to current), then P = VI. From Ohm's law, V = IR, and I = V/R. Substituting V/R for I in the power equation, P = V²/R.
For this circuit, the V = 36 V and R (the equivalent resistance) = 4.0 Ω. So, the power loss in the circuit is P = (36 V)²/(4.0 Ω) = 324 W [choice J].
