Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 5195 by sanusihammed last updated on 29/Apr/16

Commented by FilupSmith last updated on 29/Apr/16

Sadly I can′t remember how to do this.

$$\mathrm{Sadly}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{remember}\:\mathrm{how}\:\mathrm{to} \\ $$$$\mathrm{do}\:\mathrm{this}. \\ $$

Answered by prakash jain last updated on 30/Apr/16

Numbers are not quite readble, in your diagram R2 and R3 is connected in series =R_(23) =R2+R3 R4 and R5 is connected in series =R_(45) =R4+R5 R_(23) and R_(45) are in parallel (1/R_(2345) )=(1/R_(23) )+(1/R_(45) ) R_(2345) =((R_(23) R_(45) )/(R_(23) +R_(4v) )) R_(2345) is connected in series with R_1 Net Resistance=R_(eq) =R_(2345) +R_1 Current from battery=I=((12)/R_(eq) ) Current thru R_1 =I Potential difference across R_(2345) =V_1 =12−IR_1 current thru R_(23) =I_1 =(V_1 /R_(23) ) current thru R_(45) =I_2 =(V_1 /R_(45) ) current thru R_2 and R_3 =I_1 current thru R_4 and R_5 =I_2 Power =I^2 R choose I based and which resistor is 2 ohm among R_1 to R_5 .

$$\mathrm{Numbers}\:\mathrm{are}\:\mathrm{not}\:\mathrm{quite}\:\mathrm{readble},\:\mathrm{in}\:\mathrm{your}\:\mathrm{diagram} \\ $$$$ \\ $$$$\mathrm{R2}\:\mathrm{and}\:\mathrm{R3}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:=\mathrm{R}_{\mathrm{23}} =\mathrm{R2}+\mathrm{R3} \\ $$$$\mathrm{R4}\:\mathrm{and}\:\mathrm{R5}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:=\mathrm{R}_{\mathrm{45}} =\mathrm{R4}+\mathrm{R5} \\ $$$$\mathrm{R}_{\mathrm{23}} \:\mathrm{and}\:\mathrm{R}_{\mathrm{45}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{parallel} \\ $$$$\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{2345}} }=\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{23}} }+\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{45}} } \\ $$$$\mathrm{R}_{\mathrm{2345}} =\frac{\mathrm{R}_{\mathrm{23}} \mathrm{R}_{\mathrm{45}} }{\mathrm{R}_{\mathrm{23}} +\mathrm{R}_{\mathrm{4v}} } \\ $$$$\mathrm{R}_{\mathrm{2345}} \:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{R}_{\mathrm{1}} \\ $$$$\mathrm{Net}\:\mathrm{Resistance}=\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{\mathrm{2345}} +\mathrm{R}_{\mathrm{1}} \\ $$$$\mathrm{Current}\:\mathrm{from}\:\mathrm{battery}=\mathrm{I}=\frac{\mathrm{12}}{\mathrm{R}_{\mathrm{eq}} } \\ $$$$\mathrm{Current}\:\mathrm{thru}\:\mathrm{R}_{\mathrm{1}} =\mathrm{I} \\ $$$$\mathrm{Potential}\:\mathrm{difference}\:\mathrm{across}\:\mathrm{R}_{\mathrm{2345}} =\mathrm{V}_{\mathrm{1}} =\mathrm{12}−\mathrm{IR}_{\mathrm{1}} \\ $$$$\mathrm{current}\:\mathrm{thru}\:\mathrm{R}_{\mathrm{23}} =\mathrm{I}_{\mathrm{1}} =\frac{\mathrm{V}_{\mathrm{1}} }{\mathrm{R}_{\mathrm{23}} } \\ $$$$\mathrm{current}\:\mathrm{thru}\:\mathrm{R}_{\mathrm{45}} =\mathrm{I}_{\mathrm{2}} =\frac{\mathrm{V}_{\mathrm{1}} }{\mathrm{R}_{\mathrm{45}} } \\ $$$$\mathrm{current}\:\mathrm{thru}\:\mathrm{R}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{R}_{\mathrm{3}} =\mathrm{I}_{\mathrm{1}} \\ $$$$\mathrm{current}\:\mathrm{thru}\:\mathrm{R}_{\mathrm{4}} \:\mathrm{and}\:\mathrm{R}_{\mathrm{5}} =\mathrm{I}_{\mathrm{2}} \\ $$$$\mathrm{Power}\:=\mathrm{I}^{\mathrm{2}} \mathrm{R} \\ $$$$\mathrm{choose}\:\mathrm{I}\:\mathrm{based}\:\mathrm{and}\:\mathrm{which}\:\mathrm{resistor}\:\mathrm{is}\:\mathrm{2}\:\mathrm{ohm} \\ $$$$\mathrm{among}\:\mathrm{R}_{\mathrm{1}} \:\mathrm{to}\:\mathrm{R}_{\mathrm{5}} . \\ $$

Commented by prakash jain last updated on 30/Apr/16

R_1 =4,R_2 =2,R_3 =1 R_4 =5,R_5 =1 R_(23) =3 R_(45) =6 R_(2345) =((3×6)/(3+6))=2 R_(eq) =R_1 +2=6 I=2 V_1 =12−4×2=4 current thru 2 ohm (R_2 )=(4/R_(23) )=(4/3) power=I^2 R=((16)/9)×2

$$\mathrm{R}_{\mathrm{1}} =\mathrm{4},\mathrm{R}_{\mathrm{2}} =\mathrm{2},\mathrm{R}_{\mathrm{3}} =\mathrm{1} \\ $$$$\mathrm{R}_{\mathrm{4}} =\mathrm{5},\mathrm{R}_{\mathrm{5}} =\mathrm{1} \\ $$$$\mathrm{R}_{\mathrm{23}} =\mathrm{3} \\ $$$$\mathrm{R}_{\mathrm{45}} =\mathrm{6} \\ $$$$\mathrm{R}_{\mathrm{2345}} =\frac{\mathrm{3}×\mathrm{6}}{\mathrm{3}+\mathrm{6}}=\mathrm{2} \\ $$$$\mathrm{R}_{\mathrm{eq}} =\mathrm{R}_{\mathrm{1}} +\mathrm{2}=\mathrm{6} \\ $$$$\mathrm{I}=\mathrm{2} \\ $$$$\mathrm{V}_{\mathrm{1}} =\mathrm{12}−\mathrm{4}×\mathrm{2}=\mathrm{4} \\ $$$$\mathrm{current}\:\mathrm{thru}\:\mathrm{2}\:\mathrm{ohm}\:\left(\mathrm{R}_{\mathrm{2}} \right)=\frac{\mathrm{4}}{\mathrm{R}_{\mathrm{23}} }=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{power}=\mathrm{I}^{\mathrm{2}} \mathrm{R}=\frac{\mathrm{16}}{\mathrm{9}}×\mathrm{2} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com