Question

For the series-parallel arrangement shown in below Figure, find (a) the supply
current, (b) the current flowmg through each resistor and (c) the p d across each
resistor​

Answer 1

Answer:

The answer will be

a) 25 A.

b)  I through :

         R1 = 25 A;

         R2 = 6.25 A;

         R3 = 18.75A;

         R4 = 25 A;

c)P.d through :

         R1 = 62.5 V;

         R2 = 37.5V;

         R3 = 37.5 V;

         R4 = 100V;

Explanation:

Given ;

                    total volt = 200 V;

                    R1 = 2.5 ohm;   R2 = 6 ohm;R3 = 2 ohm; R4 = 4 ohm;

Let's take Rp = R2 + R3 for the purpose of understanding;

a) Total effective resistance = R1 + Rp + R4;      (i)

Here, 1/Rp = 1/R2 + 1/R3;

          1/Rp = 1/6 + 1/2;

           1/Rp = 4/6;

           Rp = 6/4 = 3/2 = 1.5 ohm;

Hence, put all the value on the equation (i);

                      =  2.5 + 1.5 + 4;

                     = 4 + 4;

                     = 8 ohm;

According to Ohm's law;

                       I = V / R;

Now put the required value;

                       I = 200 / 8;

                        = 25 A;

b) Since, R1 , Rp, R4 are in series. Hence, current will be constant;

Therefore, I through R1 = 25 A;

Now, Rp is the parallel combination of resistor R2 and R3;

If we take Rp in series (which, indeed, is in series when we take whole R3 and R2 as a single resistor) with R1 and R2, then I through Rp = 25 A

Now, p.d through Rp = I * Rp ;

                                          = 25 * 1.5;

                                          = 37.5 V;

Now, P.d. in R2 and R3 will be constant since they are in parallel combination.

Now, current through R2 = V / R2;

                                          = 37.5 / 6;

                                          = 6.25 A;

And, current through R3 = V / R3;

                                          = 37.5 / 2;

                                          = 18.75 A;

Since, R4 is also in series hence, current through R4 = 25 A;

∴ I through :

R1 = 25 A;

R2 = 6.25 A;

R3 = 18.75 A;

R4 = 25 A;

c)P.d. through R1 = I * R1;

                             = 25 * 2.5;

                            = 62.5 V;

   Here, Rp is the parallel combination of R1 and R2;

And, I in R2 = 6.25 A;

and, I in R3 = 18.75 A;

Hence, P.d. through R2 = I in R2 * R2;

                                        = 6.25 * 6;

                                        = 37.5 V.

and   P.d. through R 3 = I in R3 * R3;

                                      = 18.75 * 2;

                                      = 37.5 V.

(This shows that Potential difference in R2 and R3 is same just because they are in parallel combination.)

Now, P.d through R4 = I * R4;

                                   = 25 * 4;

                                   = 100 V;

∴ P.d through :

R1 = 62.5 V;

Rp = R2 = 37.5V;

        R3 = 37.5 V;

R4 = 100V;

That's all.

Answer 2

Answer:

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