Question

The diagram above shows three resistor connected across a cell of e.m.f 1.8V and internal resistance r. Calculate :. i) Current through 3ohm resistor. ii) The internal resistance ​

Answer 1

Answer:

i)0.6 ampere

ii)5 ohms

Explanation

i)

by ohms law,

V=IR

I=V/R

=>1.8/3

=>0.6 ampere

ii)

total resistance,

1/Rₐ = 1/R₁ + 1/R₂

=> 1/3 + 1/1.5

=>1/3 + 2/3

=>Rₐ = 1 ohm

Rₐ + R₃ = total resistance

=> 1 + 4 = 5 ohms

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Answer 2

Answer:

(i) The current through the 3-ohm resistor is 0.1 A.

(ii) The internal resistance is 1Ω.

Explanation:

(i) Current through 3 ohm

The resistors 3Ω and 1.5Ω are connected in parallel. The equivalent resistance of the combination is calculated as,

$\frac{1}{R} =\frac{1}{3} +\frac{1}{1.5}$

$\frac{1}{R}=\frac{1+2}{3}$

$\frac{1}{R} =\frac{3}{3}$

$R=1\Omega$     (1)

And the voltage across the parallel combination will be,

$V=IR$     (2)

$V=0.3\times 1$

$V=0.3 volt$     (3)

Since voltage in parallel combination remains constant. So, the current through a 3-ohm resistor by using equation (2) is given as,

$0.3=I\times 3$

$I=0.1 A$   (4)

(ii)The internal resistance

We will use the following formula,

$I=\frac{V}{R+4+r}$     (5)

Where,

I=net current through the circuit

V=net voltage of the circuit

R=resistance of the resistors in parallel combination

r=internal resistance

From the question we have,

I=0.3A

V=1.8 volt

R=1Ω        (equation (1))

By substituting the values in equation (5) we get;

$0.3=\frac{1.8}{1+4+r}$

$5+r=6$

$r=6-5$

$r=1\Omega$      (6)

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