Question

for the circuit shown in fig 4.11. Determine the equivalent resistance between A and B. Also find the current flowing from the battery​

Answer 1

Given :

• R₁ = 5Ω
• R₂ = 5Ω
• R₃ = 5Ω
• R₄ = 10Ω

Potential difference = 10 volts.

To find :

The equivalent resistance and the current flowing from the battery.

Solution :

In the given figure R₂ and R₃ are connected in series combination,

» The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. i.e.,

$\dashrightarrow\sf R_{23} = R_2 + R_3$

$\dashrightarrow\sf R_{23} = 5 + 5$

$\dashrightarrow\sf R_{23} = 10 \: ohms$

Now, R₄ and R₂₃ are connected in parallel combination,

» The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

$\dashrightarrow\sf \dfrac{1}{R_{234}} = \dfrac{1}{R_{23}} +\dfrac{1}{ R_4}$

$\dashrightarrow\sf \dfrac{1}{R_{234}} = \dfrac{1}{10} +\dfrac{1}{ 10}$

$\dashrightarrow\sf \dfrac{1}{R_{234}} = \dfrac{2}{10}$

$\dashrightarrow\sf \dfrac{1}{R_{234}} = \dfrac{1}{5}$

$\dashrightarrow\sf R_{234} = 5 \: ohms$

Now, R₁ and R₂₃₄ connected in series combination,

$\dashrightarrow\sf R_{eq} = R_1 + R_{234}$

$\dashrightarrow\sf R_{eq} = 5 + 5$

$\dashrightarrow\sf R_{eq} = 10 \: ohms$

thus, the total resistance of the circuit is 10Ω.

To find the current we can use ohm's law that is,

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula : V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

$\dashrightarrow \sf V = RI$

$\dashrightarrow \sf 10 = (10)I$

$\dashrightarrow \sf I = \dfrac{10}{10}$

$\dashrightarrow \sf I = 1 \: A$

thus, the current flowing from the battery is 1A.

Remember !

SI unit of current is ampere (A).

SI unit of resistance is ohms (Ω).