Current Electricity
Topics
- Electric current, drift velocity, mobility
- Ohm's law, resistance, V-I characteristics of ohmic and non-ohmic conductors
- Resistivity, conductivity and temperature dependence of resistance
- Electrical energy and power
- Series and parallel combination of resistors
- Internal resistance, emf, cells in series and parallel
- Kirchhoff's laws and applications
- Wheatstone bridge and metre bridge
Chapter tests
See all →Top questions
See all →- A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is : Answer: 2. 2.0 A Key idea: C and D are joined by a wire, so they are at the same potential. The circuit becomes two parallel pairs in series... 1
- A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is : Answer: 3. R/16 Key idea: cutting a wire into equal parts divides its resistance equally. Each piece has resistance R/8 . Four in parallel g... 1
- The current passing through the battery in the given circuit, is: Answer: 2. 0.5 A Key idea: the middle network is a balanced Wheatstone bridge, so the 6\ Ω resistor carries no current. Points C and D are j... 1
- Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is: Answer: 3. 2 : 9 Key idea: P=V²/R , so R∝1/P at fixed voltage. Take R_A=2 units (1 kW) and R_B=1 unit (2 kW). Series: R_s=3 , so P_s=V²/3 .... 1
- Choose the correct circuit which can achieve the bridge balance. Answer: 2. circuit (2) Key idea: A Wheatstone bridge is balanced when no current flows through G, i.e. P/Q=R/S . The top arms are 10 Ω and 1... 1
- The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is : Answer: 4. 8 V Key idea: Terminal voltage V=ε-Ir . Current I=ε/R+r=10/4+1=2 A. V=10-2×1=8 V (equivalently IR=2×4=8 V). Option 4. 1