What is the current in mA through a 500 Ω resistor that is connected to a 1.5 V battery? Show all calculations using Ohm’s law (
1 answer:
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Answer:
Explanation:
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Answer:
The option (b) is correct.
Explanation:
The expression for the power in terms of work done is as follows;
Here, W is the work done, t is the time taken and P is the power.
According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.
The expression for the equivalent resistance in the series combination is as follows;
R_{eq}=R_{1}+R_{2}......+R_{6}
Put R_{1},R_{2}......,R_{6}=R.
R_{eq}=R+R+R+R+R+R
R_{eq}=6R
It is given in the problem that a seventh resistor is added (in series).
R_{eq}=R_{1}+R_{2}......+R_{7}
Put R_{1},R_{2}......,R_{7}=R.
R_{eq}=R+R+R+R+R+R+R
R_{eq}=7R
The expression for the power in terms of voltage and resistance is as follows;
Here, R is the resistance.
From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.
If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.
Therefore, the option (b) is correct.