Question

4ohm,6ohm,8ohm are connected in parallel to a battery of 13V.
What is current through 4ohm? What is equivalent resistance

Answer 1

Answer:

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Answer 2

Given:

Three resistances 4Ω,6Ω and 8Ω are connected in parallel to a battery of 13V

To Find:

Current passing through 4Ω resistance and equivalent resistance

Solution:

We know that,

• According to Ohm's law, under certain conditions

$\orange{\underline{\boxed{\bf{V=IR}}}}$

where,

V is potential difference

I is current

R is resistance

• In parallel combination of resistances, potential difference across each resistance is same and is equal to the main potential difference i.e.

$\pink{\underline{\boxed{\bf{V=V_{1}=V_{2}=V_{3}=...=V_{n}}}}}$

• In parallel combination of resistances, equivalent resistance is given by

$\purple{\underline{\boxed{\bf{\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+.....+\dfrac{1}{R_{n}}}}}}$

$\rule{190}{1}$

Let the current passing through 4Ω resistance be I

So, on using ohm's law, we get

$\longrightarrow\rm{V=IR}$

$\longrightarrow\rm{13=I(4)}$

$\longrightarrow\rm{4I=13}$

$\longrightarrow\rm{I=\dfrac{13}{4}}$

$\longrightarrow\rm\green{I=3.25\ A}$

$\rule{190}{1}$

Now, let the equivalent resistance of given circuit be R

So,

$\longrightarrow\rm{\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}}$

$\longrightarrow\rm{\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}}$

$\longrightarrow\rm{\dfrac{1}{R}=\dfrac{6+4+3}{24}}$

$\longrightarrow\rm{\dfrac{1}{R}=\dfrac{13}{24}}$

$\longrightarrow\rm{R=\dfrac{24}{13}}$

$\longrightarrow\rm\green{R=1.85\ \Omega}$

Hence, the current through 4Ω resistance is 3.25 A and equivalent resistance of given circuit is 1.85 Ω.