**Some hard questions here -
that need a bit of detective work. **

Hints: - remember that current is the same in all series parts of a circuit

- remember that the potential difference is the same across parallel parts

- remember that the potential difference across series parts will add up to the voltage of the supply

**For each of these circuits label the
missing value of current, potential difference or resistance:**

**1. **Ammeter current = 0.3A,
Cell voltage = 9V, Resistance of bulb =

**2. **Voltage
of battery = 9V, Resistance
of buzzer = 50 Ohms, Current
= 0.05A

Resistance of the motor =

**3.**
When
switch is closed: p.d across bulb Y = 5V, Resistance of bulb Y = 400
Ohms, Resistance of bulb X = 500 Ohms,

P.d. across bulb X =

**4.**
When
switch is closed: Current flowing through circuit = 0.05A, Resistance of bulb
X = 2 kilo Ohms, P.d.
across bulb Y = 130V

Voltage
of supply =

__
__

**5.**
Current flowing out of cell = 2.5A, Current flowing into top bulb is 0.5A,
Resistance of top bulb = 10 Ohms

Resistance of lower bulb is

**Press the button to check your
answers: **