"How do changes in voltage or resistance affect current in an electric circuit?"

1 answer:

5 0

We can answer the question by looking at the Ohm's law, which gives us the relationship between voltage (V), current (I) and resistance (R) of a circuit:
$V=IR$
equivalently, we can rewrite it as
$I= \frac{V}{R}$

by looking at the equation, we can make the following observations:
1) The current is proportional to the voltage: therefore, if the voltage increases, the current increases as well; if the voltage decreases, the current decreases too.
2) The current is inversely proportional to the resistance: if the resistance increases, the current decreases, and if the resistance decreases, the current increases.

You might be interested in

What type of electromagnetic radiation can be used as a medical tracer to diagnose disease by being sent through the body's circ

Stolb23 [73]

It c.microwaves !!!!!!!!!!

4 0

2 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$