Electrical Circuits I

Fast solutions:

            Variable resistor                      Ammeter                                     Voltmeter

                    Battery                                                   Resistor                                              Bulb

                 Inductor                                 Cell                              Capacitor

1. 1/3 A

2. 1V

3. 2V

4. 2/3 W

5. 1W

Detailed solutions:

1. the total resistance of the circuit is

   $3\Omega + 6\Omega = 9\Omega$ because this is a series circuit and therefore resistances add up

   V=IR so 3 = I x 9 so I = 3/9 = 1/3 A

2. The voltage across the resistor:V=IR so V=1/3 x 3 = 1V because the current is the same through the bulb and the resistor

3. the voltage through the bulb is:
   V=IR so V=1/3 x 6 = 2V or because the total voltage across the whole circuit is 3V and the resistor is taking up 1V, we know that the bulb is taking 3V - 1V = 2V.

4. The power used by the bulb:
   Power=IV so P=1/3 x 2 = 2/3 Watts

5. The total power supplied by the cell:
   P = IV so P = 1/3 x 3 = 1 Watt
   This means that the resistor must be dissipating 1 - 2/3 = 1/3 Watts