Question

Ans the question from the attachment wd full solution.....


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Answer 1

GiveN :

• Potential Difference (V) = 9 V
• Four bulbs are connected in circuit with the resistances given as 4Ω, 2Ω, 5Ω , 7Ω

To FinD :

• Equivalent Resistance of Circuit

SolutioN :

By looking at the given figure we come to know that there are four resistors (bulbs) which are connected in a circuit to a battery of 9 V.

And also, all the resistances are in series combination. (see attached picture for simple figure of circuit)

Use formula for equivalent resistance in Series :

$\longrightarrow {\boxed{\sf{R_s \: = \: R_1 \: + \: R_2 \: + \: ..... \: R_n}}}$

Let Resistance,

• $\sf{R_1 \: = \: 2 \Omega}$
• $\sf{R_2 \: = \: 4 \Omega}$
• $\sf{R_3 \: = \: 5 \Omega}$
• $\sf{R_4 \: = \: 7 \Omega}$

Substitutes values of resistors :

$\longrightarrow \sf{R_s \: = \: R_1 \: + \: R_2 \: + \: R_3 \: + \: R_4}$

$\longrightarrow \sf{R_s \: = \: 2 \: + \: 4 \: + \: 5 \: + \: 7}$

$\longrightarrow \sf{R_s \: = \: 6 \: + \: 12}$

$\longrightarrow \sf{R_s \: = \: 18}$

$\underline{\underline{\sf{Equivalent \: Resistance \: is \: 18 \Omega}}}$

________________________

We know that in a series combination current in all the resistors remains same.

Use Ohm's Law :

$\longrightarrow \sf{V \: = \: IR}$

$\longrightarrow \sf{I \: = \: \dfrac{V}{R}}$

$\longrightarrow \sf{I \: = \: \dfrac{9}{18}}$

$\longrightarrow \sf{I \: = \: \dfrac{1}{2}}$

$\longrightarrow \sf{I \: = \: 0.5}$

$\underline{\underline{\sf{Current \: flowing \: in \: circuit \: is \: 0.5 \: A}}}$