Question

(a) With the help of a circuit diagram, deduce the equivalent resistance of two resistances connected in series.
(b) Two resistances are connected in series as shown in the diagram
(i) What is the current through the 5 ohm resistance?
(ii) What is the current through R?
(iii) What is the value of R?
(iv) What is the value of V?

Answer 1

The equivalent resistance of two resistances R1 and R2 are connected in a series manner.

Explanation:

(a) From the figure, Let the current passing through each resistor i.e., R1 and R2 be I and potential difference be V.

From the ohm’s law, V= IR ---- (i)

Since the resistors are connected in series, the current flowing with a battery of V volts are:

V1 = I x R1 ------ (ii)

V2 = I x R2 ------ (iii)

Therefore, the potential difference, V = V1 + V2 ------ (iv)

Now, from equations (i), (ii), (iii) and (iv), we have,

IR = IR1 + IR2  

So, R =R1 + R2  

Hence, the equivalent resistance of two resistances R1 and R2 connected in series as from the above figure, R = R1 + R2

(b) (i) The current through 5 ohm resistor, I = V/R = (10 V)/(5 Ω ) = 2 A.

(ii) The current through R is 2 A as in series combination, the current remains the same.

(iii)The value of R = V/I = 6/2 = 3 ohm.

The value of V = total volts flowing through the circuit, V = V1 + V2 = 10 + 6 = 16 V.

You can get all the answers of Chapter 1 in the link below:

https://brainly.in/question/14961133

Few questions related to this chapter:

The figure given below shows three resistors?

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Answer 2

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