Question

311. A current of 4aexists in a 10 resistor for
4 minute, find the charge and number
of electrons that pass through any
cross-section of the
resistor
in this time.

Answer 1

Answer:

$\boxed{ \large{ \pink{charge \: = 960 \: coulomb}}}$

$\boxed{ \green{ \large{number \: of \: electrons = 6 \times 10 ^{21} }}}$

Step-by-step explanation:

$\bold{ \underline{ \blue{Given}}} \longrightarrow \\ current \: in \: circuit = 4 \: amp \\ time \: = 4 \: minute$

$\bold{ \underline{ \red{To \: find}}} \longrightarrow \\ charge \: and \: number \: of \: electrons \: flow\\ \: through \: any \: cross \: section$

$\bold {\underline{ \orange{Concept \: used}}} \longrightarrow \\ \boxed{ \huge{ \pink{i \: = \dfrac{q}{t} }}} \\ i \: = current \\ q \: = charge \\ t \: = time \\ \boxed{ \huge{ \pink{q \: = \: n \: e}}} \\ q \: = charge \\ n \: = number \: of \: electron \\ e \: = charge \: on \: one \: electron \\ \: \: \: \: \: = 1.6 \times 10 ^{ - 19}$

$\bold{ \underline{ \blue{Solution}}} \longrightarrow \\ \: \: \: \: \: \: \: \: i \: = \dfrac{q}{t} \\ = > 4 = \dfrac{q}{4 \times 60} \\ = > q = 4 \times 240 \\ = > q = 960 \: coulomb \\ now \\ \: \: \: \: \: \: \: q \: = n \: e \\ = > 960\: = n \: \times 1.6 \times 10 ^{ - 19} \\ = > n \: = \dfrac{960}{1.6 \times 10 ^{ - 19} } \\ = > n \: = \dfrac{9600 \times {10}^{ 19} }{16} \\ = > n \: = 600 \times {10}^{ 19} \\ = > n = 6 \times {10}^{21} \\ number \: of \: electrons \: = 6 \times {10}^{21}$