5 resistor are connected in a circuit as shown. find the metre reading when circuit is closed.
Answers
Answer:
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Answer:
- The ammeter reading will be 1 Ampere
Given:
- R₁ = R₂ = R₃ = 3 Ω
- R₄ = R₅ = 0.5 Ω
- Voltage (V) = 3 V
Explanation:
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In solving such questions find the net resistance in the circuit. and then use ohm's law to find the Current through circuit.
Firstly we will consider the triangle DCE
We can see that, R₁ and R₂ are in series, let the equivalent resistance of R₁ and R₂ be R'.
Now,
⇒ R' = R₁ + R₂
Substituting the values,
⇒ R' = 3 Ω + 3 Ω
⇒ R' = 3 + 3
⇒ R' = 6
⇒ R' = 6 Ω
∴ Equivalent resistance of R₁ and R₂ is 6 Ω.
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Now, R' is parallel to R₃ resistor and let there equivalent resistance be R".
Therefore,
⇒ R" = R' R₃ / R' + R₃
∵ [1/R" = 1/R' + 1/R₃]
Substituting the values,
⇒ R" = (6 × 3) / 6 + 3
⇒ R" = 18 / 9
⇒ R" = 2
⇒ R" = 2 Ω
∴ Equivalent resistance of R' and R₃ is 2 Ω.
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Now, all the resistors i.e. R", R₄ and R₅ are connected in series. let there equivalent resistance be R.
Therefore,
⇒ R = R" + R₄ + R₅
Substituting the values,
⇒ R = 2 Ω + 0.5 Ω + 0.5 Ω
⇒ R = 2 + 0.5 + 0.5
⇒ R = 2 + 1
⇒ R = 3
⇒ R = 3 Ω
∴ Equivalent resistance of circuit is 3 Ω.
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From Ohm's law we know,
⇒ V = I R
Here,
- V Denotes Voltage.
- I Denotes current.
- R Denotes resistance.
Substituting the values,
⇒ 3 V = I × 3 Ω
⇒ 3 = I × 3
⇒ I = 3 / 3
⇒ I = 1
⇒ I = 1 A
∴ The ammeter reading will be 1 Ampere.
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