Question

A p.d. of 4 V is applied to two resistors of 6 Ω and 2 Ω connected in series. Calculate
(a) the combined resistance
(b) the current flowing
(c) the p.d. across the 6 Ω resistor

Answer 1

(a) Req = R1 + R2

= 6 + 2

= 8 ohms.

(b) V = 4V

V = iR

I = 4/8

I = 0.5A

(c) r = 6 ohms

I = 0.5A

V = iR

= 0.5 × 6

= 3V

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Answer 2

Explanation:

According to the question, the two resistors are connected in parallel and potential difference of 6 V is applied.

The combined resistance, $R=R 1+R 2=6+2=8 \text { ohm.}$.

   So, $R = 8 ohm$

The current flowing in the main circuit, I.

From ohm’s law, $V=IR$

$\mathrm{I}=\frac{V}{R}=\frac{4}{8}=0.5 \mathrm{A}.$

The potential difference across 6 ohm resistors,$V=I R$

=$=0.5 \times 6=3 \text { volts.}$