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CBSE Class 10 Science 2015 Term I Set 2

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Question : 14 of 36
Marks: +1, -0
In the given electric circuit if the current flowing through 3Ω3 \Omega resistor is 1 A1\ \mathrm{A}, find the voltage of the battery and the current II drawn from it.
Solution:  
E  =?E\;=?
I  =?I\;=?
R2  =3 ΩR_2\;=3\ \Omega
I2  =1 A (  Current flowing through   R2)I_2\;=1\ \mathrm{A}\ (\;\text{Current flowing through}\;\ R_2)
V2  =I2R2=1×3=3  volts  V_2\;=I_2 R_2=1 \times 3=3\;\text{volts}\;
Resultant resistance of R2R_2 and R3R_3 is R′R'.
  1R′  =  1R2+  1R1=  13+  16=  2+16=  36=  12\;\frac{1}{R'}\;=\;\frac{1}{R_2}+\;\frac{1}{R_1}=\;\frac{1}{3}+\;\frac{1}{6}=\;\frac{2+1}{6}=\;\frac{3}{6}=\;\frac{1}{2}
R′  =2 ΩR'\;=2\ \Omega
V′  =V2=3  volts (Voltage across   R′)V'\;= V_2=3\;\text{volts (Voltage across}\;\ R')
I′  =?I'\;=?
I′  =  V′R′=  32 A=1.5 AI'\;=\;\frac{V'}{R'}=\;\frac{3}{2}\ \mathrm{A}=1.5\ \mathrm{A}
I′  =I=1.5 AI'\;= I=1.5\ \mathrm{A}
Resultant resistance in the circuit =R=R
R  =R1+R′=2+2=4 ΩR\;= R_1+ R'=2+2=4\ \Omega
I  =  32 AI\;=\;\frac{3}{2}\ \mathrm{A}
V  =?V\;=?
V  =IR=  32×4=6  volts  V\;= IR=\;\frac{3}{2} \times 4=6\;\text{volts}\;
⇒    E  =V=6  volts  \Rightarrow\;\; E\;= V=6\;\text{volts}\;
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