Question

Two resistors r1 and r2 are connected in series and parallel. The values of the resistances are r1 = 100.0 0.1 r2 = 50.0 0.03 calculate the uncertainty in the combined resistance for both the series and the parallel arrangements.

Answer 1

Answer:

Two resistors,R1 and R2 are connected in parallel.

1/Re = 1/R1 + 1/R2

Then Re  = R1R2/R1+R2

= 50*100/50+ 100 =5000/150 = 100/3

Now, Parallel connection error:

Now, Parallel connection error:

= R21(dB) + R22(dA)/(R1 + R2(2)

= 502(3) + 1002(2)/7500 + 1002(2)/1502

= 11/9

Relative Error = (11/9)/100/3) = 0.03666

Explanation:

Answer 2

Given: The values of resistances are $R_1=(100\frac{+}{-} 0.1)ohm$ and $R_2=(50\frac{+}{-} 0.03)ohm$

To find: The uncertainty of combined resistances.

Solution:

To find the uncertainty we need to calculate the total relative error in the parallel connection.

For series connection, the total equivalent resistance will be,

$R_e_q=R_1+R_2$

$R_e_q=(100\frac{+}{-}0.1)+(50\frac{+}{-}0.03)$

$R_e_q=(150\frac{+}{-}0.13)ohm$

For parallel connection,

The equivalent resistance will be,

$R_e_q=\frac{R_1R_2}{R_1+R_2}$

⇒ $R_e_q=\frac{100*50}{100+50}$

⇒ $R_e_q=\frac{100}{3}$

The parallel connection error will be:

$=\frac{R_1^2(dB)+R_2^2(dA)}{(R_1+R_2)^2}$

$=\frac{100^2(0.1)+50^2(0.03)}{150*150}$

$=\frac{1000+75}{22500}$

$=\frac{1075}{22500}$

∴ The relative error,

$\frac{\frac{1075}{22500} }{\frac{100}{3} }$

$=0.0143$

Final answer:

The uncertainty of the connection will be 0.0143.