Question

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Answer 1

hope it's the answer to your question..

But I'm not sure!!

Answer 2

here,

Resistor BC(R1)=50ohm

Resistor BD(R2)=75ohm

Resistor AD(R3)= 50ohm

Resistor AX(R4)= 100ohm.

Now, According to the given circuit=>

Resistor(R1) and Resistor (R2) are in parallel connection.

Hence,the net resistance formed by their combination will be=>

1/(R)s=1/R1 + 1/R2

=>1/(R)s =1/50+1/100

=>1/(R)s = 3/150

=>(R)s = 50ohm ---------(i)

Now,(R)s and R3 are in parallel combination also.Hence we get=>

1/(R)g= 1/(R)s + 1/R3

=>1/(R)g = 1/30 + 1/50

=>1/(R)g = (5+3)/150

=>1/(R)g = 8/150

=>1/(R)g = 4/75

=>(R)g = 75/4ohm----------(ii)


But we know that R4 and (R)g are in series combination,hence we get the final resistance as=>

Rn = R4 + (R)g

=>Rn = 100 + 75/4

=>Rn = (400+75)/4

=>Rn = 475/4

=>Rn = 118.75 ohm


Hence,the net resistance between point X and Y will be 118.75 or 475/4 ohm.



Remember

1)Series Combination

-Net resistance is sum of all resistance in series.

-Rn= R1 + R2 + R3 +..... Rg

where,

Rn=> is the net resistance

R1,R2,R3=> are resistors in series and

Rg =>is the last resistor of the circuit.

-In these type of circuits,the net resistance is always greater than the highest value of resistance in the circuit.
-In Series combination each resistor gets same Current.
2) Parallel Combination

- in this type of connection inverse of Net resistance is the sum of inverse of all other resistance in the circuit.

-1/Rn = 1/R1 + 1/R2 + 1/R3 +....1/Rp

where,

1/Rn=> is inverse for net resistance

1/Rp is the inverse of last resistance in the circuit.

-In this type of combination the net resistance is smaller than the smallest value of resistance in the combination.
-Each if the resistor gets same potential difference when connected in parallel.