Question

For the circuit shown in given figure,determine (a) the reading on the ammeter, and (b) the value of Resistor R2

Answer 1

Answer:

the answer is explained above

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Answer 2

(I). The reading on the ammeter is 2 A.

(II). The value of second resistance is 40 Ω.

Explanation:

Given that,

First resistance $R_{1}=5\ \Omega$

Third resistance $R_{3}=20\ \Omega$

Current in first resistance = 8 A

Total current = 11 A

Let the total resistance is $R_{s}$ and the current in third resistance is I₃.

We need to calculate the potential difference

Using ohm's law

$V = I_{1}R_{1}$

Put the value into the formula

$V = 8\times5$

$V=40 V$

The voltage is same in parallel.

We need to calculate the current in third resistance

Using ohm's law

$V=I_{3}R_{3}$

$I_{3}=\dfrac{V}{R_{3}}$

Put the value into the formula

$I_{3}=\dfrac{40}{20}$

$I_{3}=2\ A$

The reading on the ammeter is 2 A.

We need to calculate the current in second resistance

Using formula of total current

$I_{s}=I_{1}+I_{2}+I_{3}$

Put the value into the formula

$11=8+I_{2}+2$

$I_{2}=1\ A$

The current in second resistance is 1 A

We need to calculate the value of second resistance

Using formula of parallel

$\dfrac{1}{R_{s}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

Put the value into the formula

$\dfrac{1}{R_{s}}=\dfrac{1}{5}+\dfrac{1}{R_{2}}+\dfrac{1}{20}$

$R_{s}= \dfrac{20R_{2}}{5R_{2}+20}$

Using ohm's law

$V=I_{c}R_{s}$

Put the value into the formula

$40=11\times\dfrac{20R_{2}}{5R_{2}+20}$

$220R_{2}-200R_{2}=800$

$R_{2}=40\ \Omega$

Hence, (I). The reading on the ammeter is 2 A.

(II). The value of second resistance is 40 Ω.

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Topic : electricity

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