Given: Two resistors in parallel, R1=0.4 Ω R_1 = 0.4\, \Omega R1=0.4Ω, R2=0.6 Ω R_2 = 0.6\, \Omega R2=0.6Ω
The formula for equivalent resistance R R R in a parallel combination is:
1R=1R1+1R2=10.4+10.6 \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{0.4} + \frac{1}{0.6} R1=R11+R21=0.41+0.61
1R=52+53=15+106=256⇒R=625=0.24 Ω \frac{1}{R} = \frac{5}{2} + \frac{5}{3} = \frac{15 + 10}{6} = \frac{25}{6} \Rightarrow R = \frac{6}{25} = 0.24\, \Omega R1=25+35=615+10=625⇒R=256=0.24Ω
Final Answer: 0·24 Ω
A battery of 6 V 6 \, \text{V} 6V is connected to the circuit as shown below. The current I I I drawn from the battery is: