Question

Topic : combination of resistors ​

Answer 1

➩Finding Effective Resistance:

(i) We can see that in case (i) ,we have just simplified the combination of resistors to get an effective resistance between point A and B.

•we can see that resistor of 10Ω(7Ω+3Ω) is connected parallel with resistor of 10Ω

$\frac{1}{Rp} = \frac{1}{R1}+ \frac{1}{R2}$

$\frac{1}{Rp} = \frac{1}{10}+ \frac{1}{10}$

$\frac{1}{Rp} = \frac{2}{10}$

$\frac{1}{Rp} = \frac{1}{5}$

$Rp = 5$

Now 5Ω is connected in series with 10Ω

5Ω+10Ω=15Ω

•Now this 15Ω resisitor gets connected in parallel with 30Ω resistor.

$\frac{1}{Rp} = \frac{1}{R1}+ \frac{1}{R2}$

$\frac{1}{Rp} = \frac{1}{15}+ \frac{1}{30}$

$\frac{1}{Rp} = \frac{2+1}{30}$

$\frac{1}{Rp} = \frac{3}{30}$

$\frac{1}{Rp} = \frac{1}{10}$

$Rp = 10Ω$

(ii) In second diagram by simplifying the diagram we obtained total 4 resistors

•we can see that resistor of 6Ω is connected parallel with resistor of 3Ω

$\frac{1}{Rp} = \frac{1}{R1}+ \frac{1}{R2}$

$\frac{1}{Rp} = \frac{1}{6}+ \frac{1}{3}$

$\frac{1}{Rp} = \frac{1+2}{6}$

$\frac{1}{Rp} = \frac{3}{6}$

$Rp= \frac{6}{3}$

$Rp= 2Ω$

Now 2Ω resistor gets connected with the resistor of 8Ω in series.

2Ω+8Ω=10Ω

Now 10Ω resistor is connected in parallel combination with resistor of 30Ω

$\frac{1}{Rp} = \frac{1}{10}+ \frac{1}{30}$

$\frac{1}{Rp} = \frac{3+1}{30}$

$\frac{1}{Rp} = \frac{4}{30}$

$Rp= \frac{30}{4}$

$Rp= \frac{15}{2}$

$Rp= 7.5Ω$