Question

a network of 4 ohm 4 ohm and 3 ohm resistance are connected to 10 volt battery.compute the equivalent resistance and current obtained from battery​

Answer 1

ohms law:

V = IR

voltage = current  X resistance.

we know that voltage = 10V so,

we must calculate the resistance now,

the first two resistors are put in a parallel position, with resistance 4 ohms and 4 ohms, so the resistance will be calculated by the parallel circuit formula.

i.e.

1/4 + 1/4 = 1/R

2/4 = 1/R

2R = 4

R = 2

next we must find the total resistance by adding the resistance value of 3 ohms to the answer,

2+3 = 5

so, the total resistance is 5 ohms.

now, to find the current, we must put the values obtained in the formula (Ohms law)

so,

V = IR

voltage = current X resistance

10 = current X 5

current = 10/5

current = 2 amp.

Answer 2

Question : -

Resistors of 4 ohm , 4 ohm and 3 ohm resistance are connected to 10 volt battery . Compute the equivalent resistance of the circuit & current obtained from the battery ?

Answer : -

Given : -

Resistors of 4 ohm , 4 ohm and 3 ohm resistance are connected to 10 volt battery .

Required to find : -

Equivalent resistance ? Current obtained from the battery ?

Formulae used : -

To find the equivalent resistance when the resistors are connected in series is ;

$\boxed{\tt{\bf{ {R}_{eq} = R_1 + R_2 + R_3 \dots \dots }}}$

To find the equivalent resistance when the resistors are connected in parallel is ;

$\boxed{\tt{\bf{ \dfrac{1}{{R}_{eq}} = \dfrac{1}{ R_1 } + \dfrac{1}{R_2} + \dfrac{1}{R_3} \dots \dots }}}$

Law used : -

Ohm's law

The voltage is directly proportional to current &resistance .

v = I R

Here,

v = voltage

I = current

R = Resistance

Solution : -

Resistors of 4 ohm , 4 ohm and 3 ohm resistance are connected to 10 volt battery .

we need to find the ;

Equivalent resistance ?

Current obtained from the battery ?

So,

From the given figure we can conclude that ;

The 2 resistors of 4 ohm are connected in parallel .

So, let's find the equivalent resistance of them . Using the formula ;

$\boxed{\tt{\bf{ \dfrac{1}{{R}_{eq}} = \dfrac{1}{ R_1 } + \dfrac{1}{R_2} + \dfrac{1}{R_3} \dots \dots }}}$

This implies ;

$\tt{ \dfrac{1}{R_{eq}} = \dfrac{ 1}{ 4 \Omega } + \dfrac{1}{ 4 \Omega } }$

$\tt{ \dfrac{1}{R_{eq} } = \dfrac{ 2 }{ 4 \Omega } }$

$\tt{ R_{eq} = \dfrac{4 \Omega}{ 2 } }$

$\tt{ R_{eq} = 2 \Omega }$

Now,

Let's find the equivalent Resistance of the whole circuit .

Since, the two 4 ohm resistors which are connected in parallel had a equivalent resistance of 2 ohm let's replace them with this resistor .

Hence,

Now, we need to apply the formula of equivalent resistance of the resistors in series ;

2 ohm & 3 ohm resistor are connected in series

$\boxed{\tt{\bf{ {R}_{eq} = R_1 + R_2 + R_3 \dots \dots }}}$

$\tt{ R_{eq} = 2 \Omega + 3 \Omega }$

$\sf{ R_{eq} = 5 \Omega }$

Hence,

Equivalent resistance of the whole circuit = 5 ohm

Now,

Let's find the current obtained from the battery .

Using the Ohm's law ;

V = IR

( Since, voltage in the circuit is 10 volt )

=> 10 = I x 5.

=> 10 = 5I

=> 5I = 10

=> I = 10/5

=> I = 2A

Hence,

Current obtained from the battery = 2A ( 2 ampere )