Electricity And Magnitism, Resistance

Electricity and Magnetism Level 1

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

0.671 A 0.671A 0.669 A 0.669A 0.679 A 0.679A 0.660 A 0.660A

Ankit Vijay by Ankit Vijay , 22, India

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2 solutions

Ankit Vijay
Jun 1, 2014

T h e r e i s n o c u r r e n t d i v i s i o n o c c u r r i n g i n a s e r i e s c i r c u i t . C u r r e n t f l o w t h r o u g h t h e c o m p o n e n t i s t h e s a m e , g i v e n b y O h m ’ s l a w a s V = I R I = V R W h e r e R , R i s t h e e q u i v a l e n t r e s i s t a n c e o f r e s i s t a n c e s 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 Ω a n d 12 Ω T h e r e s u m i s 13.4 Ω P o t e n t i a l d i f f e r e n c e , V = 9 V I = 9 13.4 = 0.671 A There\quad is\quad no\quad current\quad division\quad occurring\quad in\quad a\quad series\quad circuit.\\ \\ Current\quad flow\quad through\quad the\quad component\quad is\quad the\quad same,given\quad by\quad Ohm’s\\ \\ law\quad as\\ \\ V=IR\\ \\ I=\frac { V }{ R } \\ \\ Where\quad R,\\ \\ R\quad is\quad the\quad equivalent\quad resistance\quad of\quad resistances\quad 0.2Ω\quad ,0.3Ω\quad ,0.4Ω\quad ,0.5Ω\quad and\quad 12Ω\\ \\ There\quad sum\quad is\quad 13.4Ω\\ \\ Potential\quad difference,\quad V=9V\\ \\ I=\quad \frac { 9 }{ 13.4 } =0.671A

3 Helpful 0 Interesting 0 Brilliant 0 Confused

We can also take the ratios of Resistance. ( though I gave the answer wrong due to error in calculation :P )

Saswata Dasgupta - 6 years, 2 months ago
Bernardo Sulzbach
Jun 19, 2014

0.6716417910447761 rounds to 0.672

You truncated the number.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Never mind !

Ahmed Salah - 6 years, 8 months ago

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