Question

In the circuit given below the current through 25 ohm resistor
is

Answer 1

Answer:

Current through 25 ohm resistor is 2 A.

Explanation:

Since, the resistors are in series and parallel combinations.

So, we have find the equivalent resistance of the whole circuit.

For equivalent resistance

step 1:  40 ohm resistance is parallel with 40 ohm.

so, equivalent resistance is 20 ohm which is in series with 10 ohm.

so, equivalent resistance is 30 ohm which is in parallel with 30 ohm.

so, equivalent resistance is 15 ohm which is in series with 25 ohm.

so, equivalent resistance is 40 ohm

therefore, the total equivalent resistance of the circuit is 40 ohm.

here, voltage of the circuit is 80 V.

therefore, current flowing through the circuit is

         80 V/40 ohm = 2 A

since, 2 A current is also flowing through the 25 ohm resistor.

hence, 2 A current is flowing through the 25 ohm resistor.

 

Answer 2

Answer:

The current through 25 Ω is 20 A.

Explanation:

Here we have been given to find the current through the 25 Ω resistor.

For this we will proceed step by step in the following way:

Let us take that,

$R_{1}$ = 40 Ω

$R_{2}$ = 40 Ω

$R_{3}$ = 10 Ω

$R_{4}$ = 30 Ω

$R_{5}$ = 25 Ω

$V$ = 80 V

Here we can see that $R_{1}$ and $R_{2}$ are parallel hence we calculate their equivalent resistance to be $R_{6}$

$\frac{1}{R_{6} } = \frac{1}{R_{1} } +\frac{1}{R_{2} }$

⇒ $\frac{1}{R_{6} } = \frac{1}{40 } +\frac{1}{40 }$

⇒ $\frac{1}{R_{6} } = \frac{2}{40 }$

⇒ $\frac{1}{R_{6} } = \frac{1}{20 }$

⇒ $R_{6}$ = 20 Ω

Now we know that $R_{3}$ and $R_{6}$ are in series therefore their equivalent resistance is $R_{7}$ and its value is as below:

$R_{7} = R_{3} +R_{6}$

⇒ $R_{7} =$ 10 Ω + 20 Ω

= 30 Ω

Now we know that $R_{7}$ and $R_{4}$ are parallel therefore their equivalent resistance is $R_{8}$ which can be calculated as:

$\frac{1}{R_{8} } = \frac{1}{R_{7} } +\frac{1}{R_{4} }$

⇒ $\frac{1}{R_{8} } = \frac{1}{30 } +\frac{1}{30 }$

⇒ $\frac{1}{R_{8} } = \frac{2}{30 }$

⇒$\frac{1}{R_{8} } = \frac{1}{15 }$

⇒ $R_{8} =$ 15 Ω

And finally, we can see that $R_{8}$ and $R_{5}$ are in series and their equivalent resistance is $R_{9}$ which can be calculated as:

$R_{9} = R_{8} +R_{5}$

⇒ $R_{9} =$ 15 Ω + 25 Ω

⇒ $R_{9} =$ 40 Ω

Therefore this is the final resistance of the circuit.

From ohm's law, we know that,

$V$ = I × R

Therefore putting $V = 80$ V and R = 40 Ω we find the current as below:

$I = \frac{V}{R}$

⇒ $I = \frac{80}{40}$

⇒ $I = 20 A$

Now as we know that in series connection the current remains the same hence current flowing through 25 Ω resistor is 20 A.