if Voltage source is 12, resistance is 6800 and voltage 1 is 8 what is R2
series circuit $$R=6800 ΩU=12 VU1=8 VR2= ? Ω $$ I. I=R1U1=R2U2⇒R2=U2U1∗R1 $$ II. R=R1+R2⇒R1=R−R2 $$ III. U=U1+U2⇒U2=U−U1 $$ we substitute U2=U−U1 and R1=R−R2 $$ R2=(U−U1U1)∗(R−R2)⇒R2=(1−U1U)∗R $$ R2=(1−8 V12 V)∗6800 Ω=(1−23)∗6800 Ω=13∗6800 Ω=2266.¯6 Ω $$ R2=2266.¯6 Ω
if Voltage source is 12, resistance is 6800 and voltage 1 is 8 what is R2
series circuit $$R=6800 ΩU=12 VU1=8 VR2= ? Ω $$ I. I=R1U1=R2U2⇒R2=U2U1∗R1 $$ II. R=R1+R2⇒R1=R−R2 $$ III. U=U1+U2⇒U2=U−U1 $$ we substitute U2=U−U1 and R1=R−R2 $$ R2=(U−U1U1)∗(R−R2)⇒R2=(1−U1U)∗R $$ R2=(1−8 V12 V)∗6800 Ω=(1−23)∗6800 Ω=13∗6800 Ω=2266.¯6 Ω $$ R2=2266.¯6 Ω