Question

A 5 V battery is connected to two 20 Ω resistors which are joined together in series.
(a) Draw a circuit diagram to represent this. Add an arrow to indicate the direction of conventional current flow in the circuit.
(b) What is the effective resistance of the two resistors?
(c) Calculate the current that flows from the battery?
(d) What is the p.d. across each resistor?

Answer 1

(a) The circuit diagram is represented in attachment below.

(b) The effective resistance of the two resistors is 40 Ω.

(c) The current that flows from the battery is 0.125 A.

(d) The p.d. across each resistor is 2.5 volts.

Explanation:

Given: A 5 V battery is connected to two 20 Ω resistors which are joined together in series.

(a) The circuit diagram is represented in attachment below.

(b) Since, the two resistance are connected in series. So, the effective resistance is given by

$= > \: r1 + r2$

$= > \: 20 \: Ω \: + \: 20 \: Ω \:$

$= > \: 40 \: Ω \:$

The effective resistance of the two resistors is 40 Ω.

(c) The current that flows from the battery is given by

$v = i \times r$

$i = \frac{v}{r}$

$i = \frac{5}{40}$

$i = \frac{1}{8}$

$i = 0.125 \: amperes$

The current that flows from the battery is 0.125 A.

(d) The p.d. across each resistor is given by

$v1 = v \times \frac{r1}{r1 + r2}$

$v1 = 5 \times \frac{20}{20 + 20}$

$v1 = 2.5 \: volts$

Similarly,

$v2 = v \times \frac{r2}{r1 + r2}$

$v2 = 5 \times \frac{20}{20 + 20}$

$v2 = 2.5 \: volts$

Answer 2

Current Flowing from battery is 0.125 Amp

Explanation:

(a) The circuit diagram has been drawn, please find the same as an attachment.

(b) Effective resistance = 20 + 20 = $40\ \Omega$  

(c) As per the Ohm's law, V = IR,  

where

V is Potential Difference, I is Current and R is Resistance

Current flowing through the circuit = I = $\frac {V}{R}$ = $\frac {5}{40}$ = 0.125 amp  

(d) Potential Difference across each resistance = $I \times R$

= $0.125 \times 20$ = 2.5v