Question

A P.D. of 4 V is applied to two resistors of 6 ohms and 2 ohms connected in series. Calculate:

(a) the combined resistance

(b) the current flowing

(c) the p.d. across the 6 ohms resistor

Answer 1

Part (a):

Step 1 Given:

1. Resistance, $R_1=6\Omega$
2. Resistance, $R_2=2\Omega$

The combined resistance of the given two resistors:

The combined resistance when connected in series can be calculated by,

Step 2 : formula used,

$R_{eq}=R_1+R_2$

As resistance is in series connection.

Step 3: Solution

$$\begin{gathered} R_{eq}=6+2 \\ =8\Omega . \end{gathered}$$

Hence, the combined resistance of the given two resistors is 8 Ohms.

Part (b):

The current flowing through the circuit:

Step 1 Given

1. The current flowing $I$ can be calculated using the formula $V=IR$
2. P.D applied to two resistors of 6 ohms and 2 ohms, $V=4V$

Step 2: Formula used

$I=\frac{V}{R}$

Step 3: Solution

$$\begin{gathered} I=\frac{4}{8} \\ I=0.5A \end{gathered}$$

Hence, a current$0.5A$ is flowing through the circuit.

Part (c):

The potential difference across the 6 $\Omega$resistors:

Step 1 Given,

$I=0.5A$

$R=6\Omega$

Step 2 Formula used

$V=IR$

Step 3 Solution

$$\begin{gathered} V=0.5\times 6 \\ V=3V \end{gathered}$$

Hence, the potential difference across the 6 Ohm resistor is 3 V.