Question

Find equivalent resistance of the given circuit and current through each resistor.​

Answer 1

Answer:

Electricity is the flow of electrical power or charge. It is a secondary energy source which means that we get it from the conversion of other sources of energy, like coal, natural gas, oil, nuclear power and other natural sources, which are called primary sources.

Answer 2

Given:-

• 3 Resistance are connected in parallel

To Find :-

• We have to find the equivalent resistance of the given Circuit and the Current through each resistor .
• They are connected in parallel ,So by using formula to find the resistance in Parallel combination -

Solution:-

$\underline{\boxed{\pink{ \sf\ \ \dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+. \ .\ .\ .\ \dfrac{1}{R_n}}}}$

$\begin{gathered}\sf{We\ have\ }\begin{cases}\sf{R_1=10k\text{\O}mega}\\\sf{R_2=2k\text{\O}mega}\\\sf{R_3=1k\text{\O}mega}\end{cases}\end{gathered}$

$\begin{gathered}\dashrightarrow\sf\ \dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\\ \\ \\ \dashrightarrow\sf\ \dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{2}+\dfrac{1}{1}\\ \\ \\ \dashrightarrow\sf\ \dfrac{1}{R_p}=\dfrac{1+5+10}{10}\\ \\ \\ \dashrightarrow\sf\ \dfrac{1}{R_p}=\dfrac{16}{10}\\ \\ \\ \dashrightarrow\sf\ R_p= \cancel{\dfrac{10}{16}}\\ \\ \\ \dashrightarrow\sf\ R_p=\dfrac{5}{8} \longmapsto (0.625k\text{\O}mega)\\ \\ \sf\ 1\ kilo\text{\O}mega= 1000\text{\O}mega\\ \\ \\ \dashrightarrow\ \sf or\ \underline{\boxed{\sf\ R_p=625\text{\O}mega}}\end{gathered}$

• Now Current
• Voltage=9V
• we know that Current is inversely proportional to Resistance If resistance increases current decreases
• we know that in parallel combination Voltage remains Constant
• Using ohm's Lawp0

$\begin{gathered}\sf \Big\lgroup\ V= IR\Big\rgroup\\ \\ \\ \dashrightarrow\sf\ I_1=\dfrac{V}{R_1}\\ \\ \sf\ \ V=9V\ \ and\ R_1=\ 10k\text{\O}mega= 10000\text{\O}mega\\ \\ \\ \dashrightarrow\sf\ I_1=\dfrac{9}{10000}\\ \\ \\ \dashrightarrow\underline{\boxed{\sf Current_1= 0.0009A}}\\ \\ \dashrightarrow\sf\ I_2=\dfrac{V}{R_2}\\ \\ \sf\ \ V=9V\ \ and\ R_2=\ 2k\text{\O}mega= 2000\text{\O}mega\\ \\ \\ \dashrightarrow\sf\ I_1=\dfrac{9}{2000}\\ \\ \\ \dashrightarrow\underline{\boxed{\sf Current_2= 0.0045A}}\\ \\ \\ \dashrightarrow\sf\ I_3=\dfrac{V}{R_3}\\ \\ \sf\ \ V=9V\ \ and\ R_3=\ 1k\text{\O}mega= 1000\text{\O}mega\\ \\ \\ \dashrightarrow\sf\ I_1=\dfrac{9}{1000}\\ \\ \\ \dashrightarrow\underline{\boxed{\sf Current_3= 0.009A}}\end{gathered}$