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Neporo4naja [7]
3 years ago
12

In a series circuit, one of the resistors is replaced with a resistor having a lower resistance value. how does this affect the

current in the circuit?
Physics
2 answers:
Assoli18 [71]3 years ago
8 0
Current will increase. 
Hoochie [10]3 years ago
4 0

Answer:

The current in the circuit isn't affected i.e it remains the same

Explanation:

In a series connected circuit, same current flows across the resistors bit different voltage is passed across them. This means that no matter the value of the resistance in a series connected circuit, the current flowing through the resistors will remain the same.

Based on the above conclusion, if one of the resistors is replaced with a resistor having a lower resistance value in a series circuit, the current in the circuit will not be affected.

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Does anyone know these four?! I really need help and fast pls
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3 0
2 years ago
The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa
Sonja [21]

Answer:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

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3 years ago
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