Question

Two resistors are connected in series with a battery whose EMF is 35 V as shown in the figure below. The resistances of resistor 1 is R1=37 and of resistor 2 is R2 = 57 2. What is the value of the current flowing through the circuit? R1 E R2

Answer 1

We can see that the resistors are in series. And we know that resistors when connected in series the equivalent resistance is nothing but the algebraic sum of the resistors. Hence the equivalent resistance is given by R1 + R2= 37+57=94 Ohms.

Now using ohms law.

V=I*R

We get I=V/R=35/94 = 0.372 Amps.

2. The brightness of the bulbs directly depends upon the voltage drop across the bulbs i.e the more the voltage drop the more is the brightness hence BULB B is the brightest which means maximum voltage is consumed by bulb B followed by bulb C then A.

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