
When a Voltmeter connected across the terminals of the cell measures \[5V\] and an ammeter connected measures \[10A\]. A resistance of \[2\Omega \] is connected across the terminal of the cell. The current flowing through this resistance is:
A) \[7.5A\]
B) \[5.0A\]
C) \[2.5A\]
D) \[2.0A\]
Answer
232.8k+ views
Hint: A circuit is given with voltmeter and ammeter attached and a resistance is attached across the circuit, now to find the current \I\ flowing in the circuit we need to find the total resistance R of the unit. The formula for resistance is
\[R=\dfrac{V}{{{I}_{c}}}\]
And after that we use the current rule to find the value of the current flowing through two of the resistors as:
\[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\]
where \[{{R}_{1}}\] and \[{{R}_{2}}\] are the resistance of the circuits and \[{{I}_{c}}\] is the current in the entire circuit.
Complete step by step solution:
The measurement of the voltage of the cell passing from one connection to another is given as \[5V\].
The measurement of current using the ammeter machine is given as \[10A\].
Another resistance is added to the circuit through which the current is needed to be measured and for that we use the resistance formula to find the resistance needed.
\[\Rightarrow {{R}_{1}}=\dfrac{V}{{{I}_{c}}}\]
\[\Rightarrow {{R}_{1}}=\dfrac{5}{10}\]
\[\Rightarrow {{R}_{1}}=0.5A\]
After find the resistance of the total circuit or cell, we now have two resistances in the circuit to find the current flowing in the circuit, we use the formula of:
\[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\]
Now placing the values in the formula we get the value of the current flowing in the entire circuit or cell as:
\[\Rightarrow I=\dfrac{0.5}{0.5+{{R}_{2}}}\times 10\] With \[{{R}_{2}}=2\Omega \]
\[\Rightarrow I=\dfrac{0.5}{0.5+2}\times 10\]
\[\Rightarrow I=2A\]
Therefore, the current in the entire circuit of the cell is \[I=2A\].
Note: The formula used above is derived from Ohm's Law which states that when a current passes through a conductor is directly proportional to the current flowing through it and the formula \[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\] is taken from the current divide rule, which states that the current flowing in the circuit has parallel resistance in it and is based on Kirchhoff’s Law.
\[R=\dfrac{V}{{{I}_{c}}}\]
And after that we use the current rule to find the value of the current flowing through two of the resistors as:
\[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\]
where \[{{R}_{1}}\] and \[{{R}_{2}}\] are the resistance of the circuits and \[{{I}_{c}}\] is the current in the entire circuit.
Complete step by step solution:
The measurement of the voltage of the cell passing from one connection to another is given as \[5V\].
The measurement of current using the ammeter machine is given as \[10A\].
Another resistance is added to the circuit through which the current is needed to be measured and for that we use the resistance formula to find the resistance needed.
\[\Rightarrow {{R}_{1}}=\dfrac{V}{{{I}_{c}}}\]
\[\Rightarrow {{R}_{1}}=\dfrac{5}{10}\]
\[\Rightarrow {{R}_{1}}=0.5A\]
After find the resistance of the total circuit or cell, we now have two resistances in the circuit to find the current flowing in the circuit, we use the formula of:
\[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\]
Now placing the values in the formula we get the value of the current flowing in the entire circuit or cell as:
\[\Rightarrow I=\dfrac{0.5}{0.5+{{R}_{2}}}\times 10\] With \[{{R}_{2}}=2\Omega \]
\[\Rightarrow I=\dfrac{0.5}{0.5+2}\times 10\]
\[\Rightarrow I=2A\]
Therefore, the current in the entire circuit of the cell is \[I=2A\].
Note: The formula used above is derived from Ohm's Law which states that when a current passes through a conductor is directly proportional to the current flowing through it and the formula \[I=\dfrac{{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}{{I}_{c}}\] is taken from the current divide rule, which states that the current flowing in the circuit has parallel resistance in it and is based on Kirchhoff’s Law.
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