Advertisements
Advertisements
Question
A current of 0.8 A flows in a conductor of 40 Ω for 1 minute. The heat produced in the conductor will be ______.
Options
1445 J
1536 J
1569 J
1640 J
Advertisements
Solution
A current of 0.8 A flows in a conductor of 40 Ω for 1 minute. The heat produced in the conductor will be 1536 J.
Explanation:
From Ohm's law, we know that
V = IR
Given: I = 0.8 A, R = 40 Ω, t = 1 min (60 sec)
Now, V = I.R
V = 0.8 × 40
V = 32 volt
Power, P = V × I
P = 32 × 0.8
P = 25.6 watt
Heat = P × t
Heat = 25.6 × 60
Heat = 1536 J
RELATED QUESTIONS
The relationship between the potential difference and the current in a conductor is stated in the form of a law.
1) Name the law.
2) What does the slope of V-I graph for a conductor represent?
3) Name the material used for making the connecting wire.
The graph between V and I for a conductor is a straight line passing through the origin.
Which law is illustrated by such a graph?
A wire of resistance 9 ohm having length 30 cm is tripled on itself. What is its new resistance?
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.
State and define Ohm’s law.
The temperature of a conductor is increased. The graph best showing the variation of its resistance is:
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB.
Let the resistance of an electrical device remain constant, while the potential difference across its two ends decreases to one fourth of its initial value. What change will occur in the current through it? State the law which helps us in solving the above stated question.
The circuit depicted in the figure is employed for studying Ohm’s Law. Instead of using a standard resistor, a student opts for a glass tube filled with mercury (tube 1), connected to the circuit through two electrodes, E1 and E2. He records the readings of the ammeter and voltmeter, thereby calculating the resistance. The student repeats the experiment by substituting tube 1 with tube 2, where the same amount of mercury fills tube 2.

Neglecting internal resistance of the cell use (> or < or =) to compare:
- the resistance in both the cases.
- the voltmeter readings in both the cases.
- the specific resistance in both the cases.
