Question

three resistors 2 ohm 4 ohm and 8 ohn are connected in parallel with one another and a battery of 16 volt is connected across the combination .find current in each resistors and net current drawn from the battery ​

Answer 1

To Find :

• The Current in each Resistors.

• The Net Current drawn from the battery.

Given :

• $\bf{R_{1} = 2 \Omega}$
• $\bf{R_{2} = 4 \Omega}$

• $\bf{R_{3} = 8 \Omega}$

• $\bf{Voltage = 16 V}$

We Know :

Ohm's Law :

$\bf{\underline{\boxed{V = I \times R}}}$

EQuivalent resistance :

(When Connected in Parallel Circuit) :

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

Where ,

• $\bf{R_{n}}$ = Equivalent Resistance.

• $\bf{R_{1},R_{2},R_{3}}$ = Resistors.

Concept :

For Current In each Resistor :

We Know that in a Parallel circuit the Voltage is same .

(While in a Series Circuit the current is same)

Since , it is given that the it's connected in parallel , so the Voltage will be same for Every Resistors ,i.e,

If Voltage V is in the Resistor $R_{1}$ , then same Voltage V will be in the Resistor $R_{2}$.

For the Net Current from the battery :

According to the question , it said to find the net current From the Battery .

So first we have to find the Equivalent resistance , of the Circuit.

By using the formula for Equivalent resistance (For a Parallel Circuit) , we can find the Net or Equivalent resistance of the Circuit .

We also know that , the Voltage will be same as in the parallel circuit and after that using the ohm's Law ,we can find the net current .

Solution :

To Find the Current in Resistor (1) :

• Resistance = 2 ohm

• Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 2} \\ \\ \\ \implies \bf{\dfrac{16}{2} = I} \\ \\ \\ \implies \bf{8 A = I} \\ \\ \\ \therefore \purple{\bf{l = 8 A}}$

Hence, the current in the Resistor is 8 A.

To Find the Current in Resistor (2) :

• Resistance = 4 ohm

• Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 4} \\ \\ \\ \implies \bf{\dfrac{16}{4} = I} \\ \\ \\ \implies \bf{4 A = I} \\ \\ \\ \therefore \purple{\bf{l = 4 A}}$

Hence, the current in the Resistor is 4 A.

To Find the Current in Resistor (3) :

Resistance = 8 ohm

Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 8} \\ \\ \\ \implies \bf{\dfrac{16}{8} = I} \\ \\ \\ \implies \bf{2 A = I} \\ \\ \\ \therefore \purple{\bf{l = 2 A}}$

Hence, the current in the Resistor is 2 A.

To Find the Net Current :

EQuivalent Resistance :

Given :

• $\bf{R_{1} = 2 \Omega}$
• $\bf{R_{2} = 4 \Omega}$

• $\bf{R_{3} = 8 \Omega}$

Using the formula for Equivalent resistance in a Parallel Circuit and by substituting the values in it , we get :

$\implies \bf{\dfrac{1}{R_{n}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ..... + \dfrac{1}{R_{n}}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{4 + 2 + 1}{8}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{7}{8}} \\ \\ \\ \\ \implies \bf{R_{n} = \dfrac{8}{7}} \\ \\ \\ \\ \bf{R_{n} = 1.14(approx.)} \\ \\ \\ \\ \therefore \purple{\bf{R_{n} = 1.14 \Omega}}$

Hence , the Equivalent resistance is 8/7 ohm.

Net Current :

• $R_{n} = 1.14 \Omega$

• $V = 16 V$

Using the ohm's law and by substituting the values in it , we get :

$\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 1.14} \\ \\ \\ \implies \bf{\dfrac{16}{1.14} = I} \\ \\ \\ \implies \bf{14.02(approx.) A = I} \\ \\ \\ \therefore \purple{\bf{l = 14 A}}$

Hence , the net current is 14 A.

Answer 2

Answer:

To Find :

The Current in each Resistors.

The Net Current drawn from the battery.

Given :

$\bf \green{R_{1} = 2 \Omega}$

$\bf{R_{2} = 4 \Omega}$

$\bf{R_{3} = 8 \Omega}$

$\bf{Voltage = 16 V}$

We Know :

Ohm's Law :

$\bf \pink{\underline{\boxed{V = I \times R}}} </p><p>$

EQuivalent resistance :

(When Connected in Parallel Circuit) :

$\bf{\boxed{\underline{\dfrac{1}{R_{n}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ....... + \dfrac{1}{R_{n}}}}} </p><p></p><p>$

Where ,

$\bf{R_{n}}$

= Equivalent Resistance.

$\bf {R_{1},R_{2},R_{3}}R$

= Resistors.

Concept :

For Current In each Resistor :

We Know that in a Parallel circuit the Voltage is same .

(While in a Series Circuit the current is same)

Since , it is given that the it's connected in parallel , so the Voltage will be same for Every Resistors ,i.e,

For the Net Current from the battery :

According to the question , it said to find the net current From the Battery .

So first we have to find the Equivalent resistance , of the Circuit.

By using the formula for Equivalent resistance (For a Parallel Circuit) , we can find the Net or Equivalent resistance of the Circuit .

We also know that , the Voltage will be same as in the parallel circuit and after that using the ohm's Law ,we can find the net current .

Solution :

To Find the Current in Resistor (1) :

Resistance = 2 ohm

Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$\begin{gathered}\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 2} \\ \\ \\ \implies \bf{\dfrac{16}{2} = I} \\ \\ \\ \implies \bf{8 A = I} \\ \\ \\ \therefore \red{\bf{l = 8 A}}\end{gathered} </p><p>$

Hence, the current in the Resistor is 8 A.

To Find the Current in Resistor (2) :

Resistance = 4 ohm

Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$</p><p>\begin{gathered}\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 4} \\ \\ \\ \implies \bf{\dfrac{16}{4} = I} \\ \\ \\ \implies \bf{4 A = I} \\ \\ \\ \therefore \purple{\bf{l = 4 A}}\end{gathered} </p><p></p><p>$

Hence, the current in the Resistor is 4 A.

To Find the Current in Resistor (3) :

Resistance = 8 ohm

Voltage = 16 V

Using the ohm's law and by substituting the values in it , we get :

$\begin{gathered}\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 8} \\ \\ \\ \implies \bf{\dfrac{16}{8} = I} \\ \\ \\ \implies \bf{2 A = I} \\ \\ \\ \therefore \pink{\bf{l = 2 A}}\end{gathered} </p><p>$

Hence, the current in the Resistor is 2 A.

To Find the Net Current :

EQuivalent Resistance :

Given :

$\bf \red{R_{1} = 2 \Omega}$

$\bf \blue{R_{2} = 4 \Omega}$

$\bf \pink{R_{3} = 8 \Omega}$

Using the formula for Equivalent resistance in a Parallel Circuit and by substituting the values in it , we get :

$\begin{gathered}\implies \bf{\dfrac{1}{R_{n}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ..... + \dfrac{1}{R_{n}}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{4 + 2 + 1}{8}} \\ \\ \\ \\ \implies \bf{\dfrac{1}{R_{n}} = \dfrac{7}{8}} \\ \\ \\ \\ \implies \bf{R_{n} = \dfrac{8}{7}} \\ \\ \\ \\ \bf{R_{n} = 1.14(approx.)} \\ \\ \\ \\ \therefore \red{\bf{R_{n} = 1.14 \Omega}}\end{gathered}$

Hence , the Equivalent resistance is 8/7 ohm.

Net Current :

$R_{n} = 1.14$

Using the ohm's law and by substituting the values in it , we get :

$\begin{gathered}\implies \bf{V = I \times R} \\ \\ \\ \implies \bf{16 = I \times 1.14} \\ \\ \\ \implies \bf{\dfrac{16}{1.14} = I} \\ \\ \\ \implies \bf{14.02(approx.) A = I} \\ \\ \\ \therefore \blue{\bf{l = 14 A}}\end{gathered} </p><p>$

Hence , the net current is 14 A.