Question

3) When a resistance of 40 2 is connected in parallel with an unknown resistance, the equivalent
resistance is found to be 15 12. If the same two resistances are joined in series, their effective
resistance will be
a) 45 ohm
b) 50 ohm
c) 562 ohm
d) 64 ohm​

Answer 1

Correct Question

•When a resistance of 40 Ω is connected in parallel with an unknown resistance, the equivalent

resistance is found to be 15 Ω. If the same two resistances are joined in series, their effective

resistance will be

Given:-

• Equivalent Resistance in Parallel = 15 ohm
• Resistance of 1st Resistor ,R1 = 40 ohm

To Find:-

• Effective Resistance in Series ,Rs

Solution:-

Firstly we calculate the Resistance of 2nd Resistor .

As the 1st given condition the two Resistors are connected in parallel and their equivalent resistance is also given . Then Using Formula

• 1/Rp = 1/R1 + 1/R2

where,

• Rp denote equivalent resistance
• R1 denote resistance of 1st Resistor
• R2 denote resistance of 2nd resistor

Substitute the value we get

→ 1/15 = 1/40 + 1/R2

→ 1/15-1/40 = 1/R2

→ 1/R2 = 8-3 /120

→ 1/R2 = 5/120

→ 1/R2 = 1/24

→ R2 = 24 ohms

• Hence, Resistance of 2nd Resistor is 24 ohms.

Now, we have to calculate the effective Resistance in Series.

• Rs = R1 + R2

where,

• Rs denote effective Resistance in Series
• R1 denote resistance of 1st Resistor
• R2 denote resistance of 2nd Resistor

Substitute the value we get

→ Rs = 40 + 24

→ Rs = 64 ohms

• Hence,the effective Resistance in series is 64 ohms.
• Required Option is (D) 64 ohms.

Answer 2

CORRECT QUESTION :

When a resistance of 40Ω is connected in parallel with an unknown resistance, the equivalent resistance is found to be 15Ω. If the same two Resistances are joined in series, their effective resistance will be –

(a) 45Ω

(b) 50Ω

(c) 562Ω

(d) 64Ω

GIVEN :

• Resistance connected to parallel with an unknown resistance = 40Ω

• Joined with equivalent resistance found = 15Ω

TO FIND :

• Effective resistance = ?

STEP-BY-STEP EXPLANATION :

Now, we are asked here to find their effective resistance –

Formula used :

• 1/Rp = 1/R1 + 1/R2

[ Substituting the values as per the formula ] :

→ 1/Rp = 1/R1 + 1/R2

→ 1/15 - 1/40 = 1/R2

→ 1/R2 = 8-3/120

→ 1/R2 = 5/120

→ 1/R2 = 1/24

→ R2 = 24

Hence, R2 = 24Ω

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So, now we will have to calculate their effective resistance = ?

Formula used :

• Rs = R1 + R2

[ substituting the values as per the formula ] :

→ Rs = R1 + R2

→ Rs = 40Ω + 24Ω

→ Rs = 64Ω

Therefore, effective resistance gained = 64Ω.

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