Advertisements
Advertisements
प्रश्न
In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm AD.
Advertisements
उत्तर
Since the capacitor will act as an open circuit here, the arm CE and EF will become ineffective. The circuit is now reduced to as below
Effective resistance across AD = (BC + CD) ∥ AD
= \[\left( 3 + 3 \right) \lVert 3 = 2 \Omega\]
Total resistance of the circuit = AD + DF = 2 + 3 = 5 Ω
Total current in the circuit = \[\frac{15}{5} = 3 A\]
The current through AD and BCD will divide in the inverse ratio of the resistance.
∴ Current through AD = \[3 \times \frac{6}{\left( 6 + 3 \right)} = 3 \times \frac{6}{9} = 2 A\]
APPEARS IN
संबंधित प्रश्न
What will be the change in the current if the potential difference is kept constant and the resistance of the circuit is made four times?
- It will remain unchanged.
- It will become four times.
- It will become one-fourth.
- It will become half.
Name the law which is illustrated by the above V−I graph.
What length of copper wire of resistivity 1.7 × 10-8 Ω m and radius 1 mm is required so that its resistance is 2Ω?
A wire has a length of 2.0 m and a resistance of 5.0 Ω. Find the electric field existing inside the wire if it carries a current of 10 A.
Fig. represents the circuit used for the verification of Ohm's law. Label the parts from A to F. state the function of each.
State and define Ohm’s law.
Which of the following I-V graph represents ohmic conductors?
State Ohm's Law. Represent it mathematically.
How is electric current related to the potential difference across the terminals of a conductor?
Draw a labelled circuit diagram to verify this relationship.
