Answer
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Hint: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance. As we know that
$\therefore V = IR$
Complete answer:
The value of R
Since R and $6\Omega $ resistors are in series, same current flows through them, i.e,
Current ‘I’ through the resistor can be
$\therefore I = \dfrac{V}{R}$
Put the values
$ \Rightarrow \dfrac{{6V}}{R} = \dfrac{{12V}}{{6\Omega }}$
Simplify
$ \Rightarrow R = 3\Omega $
The reading of ammeter will be
Ammeter reading will be same as current through R,
$\therefore I = \dfrac{V}{R}$
Put the value
$ \Rightarrow I = \dfrac{{6V}}{{3\Omega }}$
$ \Rightarrow I = 2A$
The potential difference across the terminal
$\therefore Potential\,Difference = {V_1} + {V_2}$
Put the value
\[ \Rightarrow Potential\,Difference = 12 + 6\]
\[ \Rightarrow Potential\,Difference = 18V\]
Additional information:
An ammeter (from ampere meter) is a measuring instrument used to measure the current in a circuit. Electric currents are measured in amperes (A), hence the name. The ammeter is usually connected in series with the circuit in which the current is to be measured.
Note: Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit.
$\therefore V = IR$
Complete answer:
The value of R
Since R and $6\Omega $ resistors are in series, same current flows through them, i.e,
Current ‘I’ through the resistor can be
$\therefore I = \dfrac{V}{R}$
Put the values
$ \Rightarrow \dfrac{{6V}}{R} = \dfrac{{12V}}{{6\Omega }}$
Simplify
$ \Rightarrow R = 3\Omega $
The reading of ammeter will be
Ammeter reading will be same as current through R,
$\therefore I = \dfrac{V}{R}$
Put the value
$ \Rightarrow I = \dfrac{{6V}}{{3\Omega }}$
$ \Rightarrow I = 2A$
The potential difference across the terminal
$\therefore Potential\,Difference = {V_1} + {V_2}$
Put the value
\[ \Rightarrow Potential\,Difference = 12 + 6\]
\[ \Rightarrow Potential\,Difference = 18V\]
Additional information:
An ammeter (from ampere meter) is a measuring instrument used to measure the current in a circuit. Electric currents are measured in amperes (A), hence the name. The ammeter is usually connected in series with the circuit in which the current is to be measured.
Note: Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit.
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