**Solutions: C**

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

**Solutions: C**

Alternatively, you may put values to work out using the 2 basic formulae of P = IV and V = IR. Refer the video below.

Rules of Series and Parallel Circuits

To understand direct current (DC) circuits, the best way is to think in terms of river system.

Series Circuit

Parallel Circuit

Series and Parallel Circuit

Examples:

The resistance of two wires X and Y are in the ratio of 2 : 1, their lengths are in the ratio 1 : 2 and their diameters are also in the ratio of 1 : 2.

What is the ratio of the resistivities of wires X and Y?

**A** 1:2 **B** 1:1 **C** 2:1 **D** 4:1

**Solutions: Option B**

Note that when the diameter of the wire doubles, the cross-sectional area increases by 4 times.

Comparing the resistances of both X and Y as shown below, the ratio of resistivities is 1:1.

**Solutions:**

View the video on how the gauge works.

<p><a href=”https://vimeo.com/138991946″>fuel guage using ammeter and variable resistor</a> from <a href=”https://vimeo.com/user10931667″>evantoh</a> on <a href=”https://vimeo.com”>Vimeo</a>.</p>

As the fuel level drops, the float which stays on the fuel surface will descend. The rod which is attached to the float will turn clockwise about the pivot X. As the rod turns, the resistance on the variable resistor increases. This increases the resistance of the circuit. Hence the current flowing through the circuit will decreases, causing the needle to deflect more to the left, indicating towards E (empty). Thus the reading on the fuel gauge decreases.

**Solutions:**

(i) Both the fixed resistor and sensor are in series.

Total effective resistance Re = 5000 + 1000 = 6000 ohms

V = IR

12 = I x 6000

I = 0.0020 A

Hence potential across Y, V = IR (where I is constant in a series circuit)

= 0.0020 x 1000

= 2.0 V

(ii) When the temperature increases and the resistance of sensor Y decreases,** the total effective resistance of the circuit decreases**.

Since V = IR, where the R of the fixed resistor is a constant 5000 ohms, as current I increases, **the potential difference across the 5000 ohms resistor will increase.**