Q14 of 330 Page 38

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?

Here a battery of 9 V is connected in series with resistors of R1=0.2ohm, R2=0.4ohm, R3=0.3ohm, R4=0.5ohm, R5=12ohm,

So the resultant resistance = R1 + R2 + R3 + R4 + R5


R = 0.2+0.4+0.3+0.5+12=13.4ohm


As we know that


V= IR


Thus the current flow through 12ohm resistance will be


I = V/R


I = 9/13.4


I = 0.67amp.


More from this chapter

All 330 →
12

Find the current in each resistor in the circuit shown below:

 13

Explain with diagrams what is meant by the “series combination” and “parallel combination” of resistances. In which case the resultant resistance is: (i) less, and (ii) more, than either of the individual resistances?

 15

An electric bulb of resistance 20Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery. Draw the circuit diagram and calculate:

(a) total resistance of the circuit


(b) current through the circuit.


(c) potential difference across the electric bulb.


(d) potential difference across the resistance wire.

 16

Three resistors are connected as shown in the diagram.


Through the resistor 5 ohm, a current of 1 ampere is flowing


(i) What is the current through the other two resistors?


(ii) What is the p.d. across AB and across AC?


(iii) What is the total resistance?