Question

A network of resistances is constructed with 1 and 2
Q resistance as shown in figure. The 5 volt battery
between A and B has negligible internal resistance,
then current flow in the circuit is​

Answer 1

Solution :

Total Resistance in the series circuit :-

We know the formula for total resistance in a series circuit i.e,

$\boxed{\bf{R_{e} = R_{1} + R_{2} + R_{3} + ... + R_{n}}}$

Where :

• Re = Equivalent resistance.

• R = Resistance in each resistor.

Now by using the formula for Equivalent resistance and substituting the values in it, we get :

$:\implies \bf{R_{e} = R_{1} + R_{2} + R_{3} + ... + R_{n}}$

$:\implies \bf{R_{e} = AG + HG + HF}$

$:\implies \bf{R_{e} = 1 + 1 + 1}$

$:\implies \bf{R_{e} = 3}$

$\boxed{\therefore \bf{R_{e} = 3\:\Omega}}$

Hence the total resistance in the series combination is 3 Ω.

Total Resistance in the Parallel Circuit :-

We know the formula for total resistance in a Parallel Circuit i.e,

$\boxed{\bf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}}$

Where :

• Re = Equivalent resistance.

• R = Resistance in each resistor.

Now by using the formula for Equivalent resistance and substituting the values in it, we get :

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1}{EF} + \dfrac{1}{GC} + \dfrac{1}{HD} + \dfrac{1}{AF}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{3}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1 + 1 + 1}{2} + \dfrac{1}{3}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{3}{2} + \dfrac{1}{3}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{9 + 2}{6}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{11}{6}}$

$:\implies \bf{R_{e} = \dfrac{6}{11}}$

$:\implies \bf{R_{e} = 0.55}$

$\boxed{\therefore \bf{R_{e} = 0.6\:\Omega}}$

Hence the total Resistance in the parallel combination is 0.6 Ω.

Current flowing in the circuit :

We know the ohm's law i.e,

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

Where :

• V = Potential Difference.

• I = Current.

• R = Resistance.

By using the ohm's law and substituting the values in it, we get :

$:\implies \bf{V = IR}$

$:\implies \bf{5 = I \times 0.6}$

$:\implies \bf{\dfrac{5}{0.6} = I}$

$:\implies \bf{8.333 = I}$

$:\implies \bf{I = 8.4}$

$\boxed{\therefore \bf{I = 8.4\:A}}$

Hence the current flowing in the circuit is 8.4 A.