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

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 .

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

According to Kirchoff's voltage rule, the algebraic sum of all voltage around any closed loop in a circuit is equal to zero. Let us label the circuit with currents in accordance with Kirchoff's rule. Apply Kirchoff's voltage rule in the loop 1 4 - 2 I 1 - 8 I 1 + I 2 - 6 I 1 = 0 4 - 16 I 1 - 8 I 2 = 0 … ( i ) Apply Kirchoff's voltage rule in the loop 2 6 - 4 I 2 - 8 I 1 + I 2 - 1 I 2 = 0 4 - 8 I 1 - 13 I 2 = 0 ⋯ ( ii ) By solving equations (i) and (ii) We get, I 1 = 5 36 A , I 2 = 2 9 A So, the current flows through the 8 Ω is I 1 + I 2 = 5 36 + 2 9 = 13 36 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

E M B I B E

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

MEDIUM

12th Meghalaya Board

IMPORTANT

Earn 100

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

50% studentsanswered this correctly

Important Questions on Current Electricity

EASY

12th Meghalaya Board

IMPORTANT

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

State Kirchhoff’s voltage law.

MEDIUM

12th Meghalaya Board

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?

MEDIUM

12th Meghalaya Board

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.

Question Image