Question

two resistors are connected in series at zero degree celsius the value of first and second resistor are respectively R and 2r the temperature coefficient of resistance of first and second resistor are alpha and 2 alpha respectively the equivalent temperature coefficient of the resistor for the combination is

Answer 1

Solution :

⏭ Given:

✏ Two resistors are connected in series.

✏ Resistance of 1st resistor = R

✏ Resistance of 2nd resistor = 2R

✏ Temp. co-efficient of R1 = α

✏ Temp. co-efficient of R2 = 2α

⏭ To Find:

✏ Equivalent temp. co-efficient of the equivalent resistor for the series connection.

⏭ Formula:

✏ Change in resistance due to change in temp. is given by

• ΔR = Ro(αΔT)
• R'-Ro = Ro(αΔT)
• R' = Ro(1 + αΔT)....(☆)

⏭ Terms indication:

• R' denotes resistance at T°C
• Ro denotes resistance at 0°C
• α denotes temp. co-efficient
• ΔT denoted change in temp.

⏭ Calculation:

✏ Equivalent resistance at 0°C

• Ro = R1 + R2
• Ro = R + 2R
• Ro = 3R

✏ Equivalent resistance at T°C

• R' = R'1 + R'2

R'1 = R(1 + αΔT)

R'2 = 2R(1 + 2αΔT)

• R' = R + RαΔT + 2R + 4RαΔT
• R' = 3R + 5RαΔT

✏ Putting all values in (☆) equation

• 3R + 5RαΔT = 3R[1 + (α)sΔT]
• 3R + 5RαΔT = 3R + 3R(α)sΔT
• 5RαΔT = 3R(α)sΔT
• 5α = 3(α)s

✏ (α)s = 5α/3

Answer 2

Answer:

Given:

2 resistors have been connected in Series. The value of resistances are R and 2R at 0°C. The coefficient of resistance are α and 2α respectively.

To find:

Equivalent coefficient of resistance.

Concept:

Since the resistors have been placed in series , we will consider the combination of resistances to be a single resistance with equivalent temperature coefficient.

Calculation:

Let the equivalent coefficient be k

$R \: eq. = R1 + R2$

$= > 3R(1 + k\Delta \theta) = R \{1 + \alpha \Delta \theta \} + 2R \{1 + (2 \alpha )\Delta \theta \}$

Cancelling R term :

$= > 3k\Delta \theta = \alpha \Delta \theta + 2(2 \alpha )\Delta \theta$

$= > 3k = \alpha + 4 \alpha$

$= > 3k = 5 \alpha$

$= > k = \dfrac{5 \alpha }{3}$

So final answer :

$\boxed{ \red{ \bold{ \huge{ k = \dfrac{5 \alpha }{3} }}}}$