Question

Two resistors are connected in series connection with a 10 volt battery. Resistance of each resistor is 5 ohm. Calculate the charges transferred in 10 seconds.

Answer 1

$\Large{\underbrace{\underline{\sf{Understanding\; the\: Concept}}}}$

Here in this question, concept of ohm's law is used. We are given a circuit of 2 resistors connected in series with a 10 volt battery. We are asked to find the number of charges transferred in 10 Seconds. Firstly we have to find the equivalent resistance of the circuit and then by Applying ohm's law, we will obtain the value of current and atlast, by applying formula for current we will obtain the number of charges transferred.

So let's do it!!

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Given:-

• Value of each resistor=5 Ω

• Potential of battery,V=10 v

• Time,t=10 seconds

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To find:-

• No. Of charges,Q

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Solution:-

Resistors in series can be expressed as::

⇒ R=R₁+R₂+R₃...

By substituting values of resistors::

⇒ R=5Ω+5Ω

⇒ R=10Ω

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Now applying ohm's law for finding current::

⇒ V=IR

⇒ 10v=I(10Ω)

⇒ 10v/10Ω=I

⇒ 1 Ampere= I

We have obtained the value of current.

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Now Applying formula for current::

⇒ I=Q/t

⇒ 1 Ampere=Q/10seconds

⇒ (1 Amp)(10 sec)=Q

⇒ 10 c=Q

So the required value of charge is 10 coulomb.

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