Find the current i in the circuit given below: Given, forward resistance of the diode is r f = 1 Ω , R = 2 Ω and V = 10 v o l t s

Here, in the given circuit D 2 will be in reverse bias so it will act as open circuit and the diodes D 1 and D 3 will be in forward bias. By applying KVL, V = i r f + i r f + i R ⇒ 10 V = i ( 1 Ω + 1 Ω + 2 Ω ) ⇒ i = 10 V 4 Ω = 2 . 5 A Hence, the current in the circuit is 2 . 5 A .

Find the current i in the circuit given below: Given, forward resistance of the diode is r f = 1 Ω , R = 2 Ω and V = 10 v o l t s

Here, in the given circuit D 2 will be in reverse bias so it will act as open circuit and the diodes D 1 and D 3 will be in forward bias. By applying KVL, V = i r f + i r f + i R ⇒ 10 V = i ( 1 Ω + 1 Ω + 2 Ω ) ⇒ i = 10 V 4 Ω = 2 . 5 A Hence, the current in the circuit is 2 . 5 A .

Find I in the following circuit : [Assume negligible internal resistance of the cell.]

Rearranging the resistances the given circuit will form a Wheatstone's bridge as shown in the figure. Now, the condition of the Wheatstone bridge is satisfied as P Q = R S where, P , Q , R , S are the sides of the bridge. From the above, the diagram 2 2 = 6 6 , satisfies the condition for the wheat stone bridge. Therefore, no current is followed through the central resistance. So, neglecting the middle resistance. Now, the total resistance of the upper side is 4 Ω as they are connected in series. Similarly, the total resistance of the lower side is 12 Ω . Hence, the equivalent resistance can be given as, 1 R = 1 4 + 1 12 = 3 + 1 12 R = 3 Ω So, the value of the current I = V R I = 3 × 6 = 18 A

Why do we prefer potentiometer to compare e.m.fs of cells than a voltmeter?

The potentiometer is preferred over voltmeter when there is a measurement of emf of the cell because the potentiometer does not draw any current as it is a null device. Whereas the voltmeter draws current from the cell. This is the reason why a potentiometer is preferred over voltmeter in order to get the actual measurement of e.m.f.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2020>Q 17

HARD

12th Meghalaya Board

IMPORTANT

Earn 100

From the following circuit, using Kirchoff's rule, find the current that flows through the $8 Ω$ resistor :

Important Questions on Current Electricity

MEDIUM

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IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2018>Q 1

Which of the following quantities does not change when a resistor is heated by passing current through it?

MEDIUM

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IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2018>Q 2

If in the following circuit, power dissipated is $150 W$ , then $R$ is

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EASY

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IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2018>Q 3

State Kirchhoff’s voltage law.

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IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2018>Q 4

The potential difference across a given wire is increased. What happens to the drift velocity of the free electrons?

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IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Meghalaya-2018>Q 6

The circuit shown in the figure below contains a battery $B$ , a rheostat $Rh$ and identical lamps $P$ and $Q$ . What will happen to the brightness of the lamps if the resistance through the rheostat is increased? Give reasons.

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