Question

18. In the circuit shown, R, = 412
R = R3 = 15 S2, R = 30 12 and
E = 10 V. Calculate the
equivalent resistance of the
circuit and the current in
each resistor.
​

Answer 1

Answer:

u412

Explanation:

Answer 2

Answer:

Given, R1 = 4 Ω, R2 = R3 = 15 Ω, R4 = 30 Ω and E = 10 V

Here, R2 , R3 and R4 are connected parallel to each other.

Therefore, equivalent resistance is given by,

1 over straight R equals straight space 1 over straight R subscript 2 plus 1 over straight R subscript 3 plus 1 over straight R subscript 4

space space space space space equals straight space 1 over 15 plus 1 over 15 plus 1 over 30

rightwards double arrow straight space straight R straight space equals straight space 6 straight capital omega space is space the space effective space resistance.

Now, R1 is in series with R. So, equivalent resistance is given by,

R subscript e q end subscript space equals space R space plus R subscript 1

italic space italic space italic space italic space italic space italic space italic equals italic space italic 6 italic plus italic space italic 4 italic space italic space

italic space italic space italic space italic space italic space italic space equals space 10 space capital omega

Current I1 is given by, I subscript 1 space equals space E over R subscript e q end subscript equals 10 over 10 equals space 1 space A ... (1)

This current is divided at A into three parts I2, I3, I4 .

therefore space I subscript 2 plus I subscript 3 plus I subscript 4 space equals space 1 ... (2)

Also,

A l s o italic comma italic space

I subscript 2 R subscript 2 space equals space I subscript 3 R subscript 3 space equals space I subscript 4 R subscript 4

rightwards double arrow space I subscript 2 space cross times 15 equals I subscript 3 space cross times 15 space equals I subscript 4 space cross times 30

rightwards double arrow space I subscript 2 equals I subscript 3 equals 2 I subscript 4 italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic. italic.. space left parenthesis 3 right parenthesis

Now, putting the values of I subscript 2 and straight I subscript 3 in (2), we get

Error converting from MathML to accessible text.

Thus, I1 = 1A, I2 = I3 = 0.4 A and I4 = 0.2 A.